Volume 1
Calculus Volume 1
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Table of Contents
Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Chapter 1: Functions and Graphs . . . . . . . . . . . . . . . . .
1.1 Review of Functions . . . . . . . . . . . . . . . . . . . .
1.2 Basic Classes of Functions . . . . . . . . . . . . . . . .
1.3 Trigonometric Functions . . . . . . . . . . . . . . . . . .
1.4 Inverse Functions . . . . . . . . . . . . . . . . . . . . .
1.5 Exponential and Logarithmic Functions . . . . . . . . . .
Chapter 2: Limits . . . . . . . . . . . . . . . . . . . . . . . . . .
2.1 A Preview of Calculus . . . . . . . . . . . . . . . . . . .
2.2 The Limit of a Function . . . . . . . . . . . . . . . . . . .
2.3 The Limit Laws . . . . . . . . . . . . . . . . . . . . . . .
2.4 Continuity . . . . . . . . . . . . . . . . . . . . . . . . .
2.5 The Precise Definition of a Limit . . . . . . . . . . . . . .
Chapter 3: Derivatives . . . . . . . . . . . . . . . . . . . . . . .
3.1 Defining the Derivative . . . . . . . . . . . . . . . . . . .
3.2 The Derivative as a Function . . . . . . . . . . . . . . . .
3.3 Differentiation Rules . . . . . . . . . . . . . . . . . . . .
3.4 Derivatives as Rates of Change . . . . . . . . . . . . . .
3.5 Derivatives of Trigonometric Functions . . . . . . . . . .
3.6 The Chain Rule . . . . . . . . . . . . . . . . . . . . . .
3.7 Derivatives of Inverse Functions . . . . . . . . . . . . . .
3.8 Implicit Differentiation . . . . . . . . . . . . . . . . . . .
3.9 Derivatives of Exponential and Logarithmic Functions . . .
Chapter 4: Applications of Derivatives . . . . . . . . . . . . . .
4.1 Related Rates . . . . . . . . . . . . . . . . . . . . . . .
4.2 Linear Approximations and Differentials . . . . . . . . . .
4.3 Maxima and Minima . . . . . . . . . . . . . . . . . . . .
4.4 The Mean Value Theorem . . . . . . . . . . . . . . . . .
4.5 Derivatives and the Shape of a Graph . . . . . . . . . . .
4.6 Limits at Infinity and Asymptotes . . . . . . . . . . . . . .
4.7 Applied Optimization Problems . . . . . . . . . . . . . .
4.8 L’Hôpital’s Rule . . . . . . . . . . . . . . . . . . . . . . .
4.9 Newton’s Method . . . . . . . . . . . . . . . . . . . . . .
4.10 Antiderivatives . . . . . . . . . . . . . . . . . . . . . .
Chapter 5: Integration . . . . . . . . . . . . . . . . . . . . . . .
5.1 Approximating Areas . . . . . . . . . . . . . . . . . . . .
5.2 The Definite Integral . . . . . . . . . . . . . . . . . . . .
5.3 The Fundamental Theorem of Calculus . . . . . . . . . .
5.4 Integration Formulas and the Net Change Theorem . . . .
5.5 Substitution . . . . . . . . . . . . . . . . . . . . . . . . .
5.6 Integrals Involving Exponential and Logarithmic Functions
5.7 Integrals Resulting in Inverse Trigonometric Functions . .
Chapter 6: Applications of Integration . . . . . . . . . . . . . .
6.1 Areas between Curves . . . . . . . . . . . . . . . . . . .
6.2 Determining Volumes by Slicing . . . . . . . . . . . . . .
6.3 Volumes of Revolution: Cylindrical Shells . . . . . . . . .
6.4 Arc Length of a Curve and Surface Area . . . . . . . . .
6.5 Physical Applications . . . . . . . . . . . . . . . . . . . .
6.6 Moments and Centers of Mass . . . . . . . . . . . . . .
6.7 Integrals, Exponential Functions, and Logarithms . . . . .
6.8 Exponential Growth and Decay . . . . . . . . . . . . . .
6.9 Calculus of the Hyperbolic Functions . . . . . . . . . . .
Appendix A: Table of Integrals . . . . . . . . . . . . . . . . . . .
Appendix B: Table of Derivatives . . . . . . . . . . . . . . . . .
Appendix C: Review of Pre-Calculus . . . . . . . . . . . . . . .
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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This OpenStax book is available for free at http://cnx.org/content/col11964/1.2
Preface
1
PREFACE
Welcome to Calculus Volume 1, an OpenStax resource. This textbook was written to increase student access to high-quality
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Format
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About Calculus Volume 1
C alculus is designed for the typical two- or three-semester general calculus course, incorporating innovative features to
enhance student learning. The book guides students through the core concepts of calculus and helps them understand
how those concepts apply to their lives and the world around them. Due to the comprehensive nature of the material, we
are offering the book in three volumes for flexibility and efficiency. Volume 1 covers functions, limits, derivatives, and
integration.
Coverage and Scope
Our Calculus Volume 1 textbook adheres to the scope and sequence of most general calculus courses nationwide. We have
worked to make calculus interesting and accessible to students while maintaining the mathematical rigor inherent in the
subject. With this objective in mind, the content of the three volumes of Calculus have been developed and arranged to
provide a logical progression from fundamental to more advanced concepts, building upon what students have already
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enable students not just to recognize concepts, but work with them in ways that will be useful in later courses and future
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dedicated to the project.
Volume 1
Chapter 1: Functions and Graphs
2
Preface
Chapter 2: Limits
Chapter 3: Derivatives
Chapter 4: Applications of Derivatives
Chapter 5: Integration
Chapter 6: Applications of Integration
Volume 2
Chapter 1: Integration
Chapter 2: Applications of Integration
Chapter 3: Techniques of Integration
Chapter 4: Introduction to Differential Equations
Chapter 5: Sequences and Series
Chapter 6: Power Series
Chapter 7: Parametric Equations and Polar Coordinates
Volume 3
Chapter 1: Parametric Equations and Polar Coordinates
Chapter 2: Vectors in Space
Chapter 3: Vector-Valued Functions
Chapter 4: Differentiation of Functions of Several Variables
Chapter 5: Multiple Integration
Chapter 6: Vector Calculus
Chapter 7: Second-Order Differential Equations
Pedagogical Foundation
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modern applications and methods. Derivations and explanations are based on years of classroom experience on the part
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Motivational applications cover important topics in probability, biology, ecology, business, and economics, as well as areas
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explore interesting sidelights in pure and applied mathematics, from determining a safe distance between the grandstand and
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of a skydiver. Chapter Opening Applications pose problems that are solved later in the chapter, using the ideas covered in
that chapter. Problems include the hydraulic force against the Hoover Dam, and the comparison of relative intensity of two
earthquakes. Definitions, Rules, and Theorems are highlighted throughout the text, including over 60 Proofs of theorems.
Assessments That Reinforce Key Concepts
In-chapter Examples walk students through problems by posing a question, stepping out a solution, and then asking students
to practice the skill with a “Checkpoint” question. The book also includes assessments at the end of each chapter so
students can apply what they’ve learned through practice problems. Many exercises are marked with a [T] to indicate they
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chapter so students can apply what they’ve learned through practice problems.
Early or Late Transcendentals
Calculus Volume 1 is designed to accommodate both Early and Late Transcendental approaches to calculus. Exponential
and logarithmic functions are introduced informally in Chapter 1 and presented in more rigorous terms in Chapter 6.
Differentiation and integration of these functions is covered in Chapters 3–5 for instructors who want to include them with
other types of functions. These discussions, however, are in separate sections that can be skipped for instructors who prefer
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Comprehensive Art Program
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This OpenStax book is available for free at http://cnx.org/content/col11964/1.2
Preface
3
Additional Resources
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About the Authors
Senior Contributing Authors
Gilbert Strang, Massachusetts Institute of Technology
Dr. Strang received his PhD from UCLA in 1959 and has been teaching mathematics at MIT ever since. His Calculus online
textbook is one of eleven that he has published and is the basis from which our final product has been derived and updated
for today’s student. Strang is a decorated mathematician and past Rhodes Scholar at Oxford University.
Edwin “Jed” Herman, University of Wisconsin-Stevens Point
Dr. Herman earned a BS in Mathematics from Harvey Mudd College in 1985, an MA in Mathematics from UCLA in
1987, and a PhD in Mathematics from the University of Oregon in 1997. He is currently a Professor at the University of
Wisconsin-Stevens Point. He has more than 20 years of experience teaching college mathematics, is a student research
mentor, is experienced in course development/design, and is also an avid board game designer and player.
Contributing Authors
Catherine Abbott, Keuka College
4
Nicoleta Virginia Bila, Fayetteville State University
Sheri J. Boyd, Rollins College
Joyati Debnath, Winona State University
Valeree Falduto, Palm Beach State College
Joseph Lakey, New Mexico State University
Julie Levandosky, Framingham State University
David McCune, William Jewell College
Michelle Merriweather, Bronxville High School
Kirsten R. Messer, Colorado State University - Pueblo
Alfred K. Mulzet, Florida State College at Jacksonville
William Radulovich (retired), Florida State College at Jacksonville
Erica M. Rutter, Arizona State University
David Smith, University of the Virgin Islands
Elaine A. Terry, Saint Joseph’s University
David Torain, Hampton University
Reviewers
Marwan A. Abu-Sawwa, Florida State College at Jacksonville
Kenneth J. Bernard, Virginia State University
John Beyers, University of Maryland
Charles Buehrle, Franklin & Marshall College
Matthew Cathey, Wofford College
Michael Cohen, Hofstra University
William DeSalazar, Broward County School System
Murray Eisenberg, University of Massachusetts Amherst
Kristyanna Erickson, Cecil College
Tiernan Fogarty, Oregon Institute of Technology
David French, Tidewater Community College
Marilyn Gloyer, Virginia Commonwealth University
Shawna Haider, Salt Lake Community College
Lance Hemlow, Raritan Valley Community College
Jerry Jared, The Blue Ridge School
Peter Jipsen, Chapman University
David Johnson, Lehigh University
M.R. Khadivi, Jackson State University
Robert J. Krueger, Concordia University
Tor A. Kwembe, Jackson State University
Jean-Marie Magnier, Springfield Technical Community College
Cheryl Chute Miller, SUNY Potsdam
Bagisa Mukherjee, Penn State University, Worthington Scranton Campus
Kasso Okoudjou, University of Maryland College Park
Peter Olszewski, Penn State Erie, The Behrend College
Steven Purtee, Valencia College
Alice Ramos, Bethel College
Doug Shaw, University of Northern Iowa
Hussain Elalaoui-Talibi, Tuskegee University
Jeffrey Taub, Maine Maritime Academy
William Thistleton, SUNY Polytechnic Institute
A. David Trubatch, Montclair State University
Carmen Wright, Jackson State University
Zhenbu Zhang, Jackson State University
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Preface
Chapter 1 | Functions and Graphs
5
1 | FUNCTIONS AND
GRAPHS
Figure 1.1 A portion of the San Andreas Fault in California. Major faults like this are the sites of most of the strongest
earthquakes ever recorded. (credit: modification of work by Robb Hannawacker, NPS)
Chapter Outline
1.1 Review of Functions
1.2 Basic Classes of Functions
1.3 Trigonometric Functions
1.4 Inverse Functions
1.5 Exponential and Logarithmic Functions
Introduction
In the past few years, major earthquakes have occurred in several countries around the world. In January 2010, an
earthquake of magnitude 7.3 hit Haiti. A magnitude 9 earthquake shook northeastern Japan in March 2011. In April 2014,
an 8.2-magnitude earthquake struck off the coast of northern Chile. What do these numbers mean? In particular, how
does a magnitude 9 earthquake compare with an earthquake of magnitude 8.2? Or 7.3? Later in this chapter, we show
how logarithmic functions are used to compare the relative intensity of two earthquakes based on the magnitude of each
earthquake (see Example 1.39).
Calculus is the mathematics that describes changes in functions. In this chapter, we review all the functions necessary
to study calculus. We define polynomial, rational, trigonometric, exponential, and logarithmic functions. We review how
to evaluate these functions, and we show the properties of their graphs. We provide examples of equations with terms
involving these functions and illustrate the algebraic techniques necessary to solve them. In short, this chapter provides the
foundation for the material to come. It is essential to be familiar and comfortable with these ideas before proceeding to the
formal introduction of calculus in the next chapter.
6
Chapter 1 | Functions and Graphs
1.1 | Review of Functions
Learning Objectives
1.1.1
1.1.2
1.1.3
1.1.4
1.1.5
1.1.6
1.1.7
Use functional notation to evaluate a function.
Determine the domain and range of a function.
Draw the graph of a function.
Find the zeros of a function.
Recognize a function from a table of values.
Make new functions from two or more given functions.
Describe the symmetry properties of a function.
In this section, we provide a formal definition of a function and examine several ways in which functions are
represented—namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. We
also define composition of functions and symmetry properties. Most of this material will be a review for you, but it serves
as a handy reference to remind you of some of the algebraic techniques useful for working with functions.
Functions
Given two sets A and B, a set with elements that are ordered pairs (x, y), where x is an element of A and y is an
element of B, is a relation from A to B. A relation from A to B defines a relationship between those two sets. A
function is a special type of relation in which each element of the first set is related to exactly one element of the second
set. The element of the first set is called the input; the element of the second set is called the output. Functions are used all
the time in mathematics to describe relationships between two sets. For any function, when we know the input, the output is
determined, so we say that the output is a function of the input. For example, the area of a square is determined by its side
length, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in the
air can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function of
the weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology to
study them.
Definition
A function f consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output. The
set of inputs is called the domain of the function. The set of outputs is called the range of the function.
For example, consider the function f , where the domain is the set of all real numbers and the rule is to square the input.
Then, the input x = 3 is assigned to the output 3 2 = 9. Since every nonnegative real number has a real-value square root,
every nonnegative number is an element of the range of this function. Since there is no real number with a square that is
negative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative real
numbers.
For a general function f with domain D, we often use x to denote the input and y to denote the output associated with
x. When doing so, we refer to x as the independent variable and y as the dependent variable, because it depends on x.
Using function notation, we write y = f (x), and we read this equation as “y equals f of x.” For the squaring function
described earlier, we write f (x) = x 2.
The concept of a function can be visualized using Figure 1.2, Figure 1.3, and Figure 1.4.
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Chapter 1 | Functions and Graphs
7
Figure 1.2 A function can be visualized as an input/output
device.
Figure 1.3 A function maps every element in the domain to
exactly one element in the range. Although each input can be
sent to only one output, two different inputs can be sent to the
same output.
Figure 1.4 In this case, a graph of a function f has a domain
of {1, 2, 3} and a range of {1, 2}. The independent variable
is x and the dependent variable is y.
Visit this applet link (http://www.openstaxcollege.org/l/grapherrors) to see more about graphs of
functions.
We can also visualize a function by plotting points (x, y) in the coordinate plane where y = f (x). The graph of a function
is the set of all these points. For example, consider the function f , where the domain is the set D = {1, 2, 3} and the
rule is f (x) = 3 − x. In Figure 1.5, we plot a graph of this function.
8
Chapter 1 | Functions and Graphs
Figure 1.5 Here we see a graph of the function f with
domain {1, 2, 3} and rule f (x) = 3 − x. The graph consists
of the points (x, f (x)) for all x in the domain.
Every function has a domain. However, sometimes a function is described by an equation, as in f (x) = x 2, with no
specific domain given. In this case, the domain is taken to be the set of all real numbers x for which f (x) is a real number.
For example, since any real number can be squared, if no other domain is specified, we consider the domain of f (x) = x 2
to be the set of all real numbers. On the other hand, the square root function f (x) = x only gives a real output if x is
nonnegative. Therefore, the domain of the function f (x) = x is the set of nonnegative real numbers, sometimes called the
natural domain.
For the functions f (x) = x 2 and f (x) = x, the domains are sets with an infinite number of elements. Clearly we cannot
list all these elements. When describing a set with an infinite number of elements, it is often helpful to use set-builder or
interval notation. When using set-builder notation to describe a subset of all real numbers, denoted ℝ, we write
⎧
⎨
⎩
x|x has some property .
⎫
⎬
⎭
We read this as the set of real numbers x such that x has some property. For example, if we were interested in the set of
real numbers that are greater than one but less than five, we could denote this set using set-builder notation by writing
{x|1 < x < 5}.
A set such as this, which contains all numbers greater than a and less than b, can also be denoted using the interval
notation (a, b). Therefore,
(1, 5) = x|1 < x < 5 .
⎧
⎨
⎩
⎫
⎬
⎭
The numbers 1 and 5 are called the endpoints of this set. If we want to consider the set that includes the endpoints, we
would denote this set by writing
[1, 5] = {x|1 ≤ x ≤ 5}.
We can use similar notation if we want to include one of the endpoints, but not the other. To denote the set of nonnegative
real numbers, we would use the set-builder notation
{x|0 ≤ x}.
The smallest number in this set is zero, but this set does not have a largest number. Using interval notation, we would use
the symbol ∞, which refers to positive infinity, and we would write the set as
[0, ∞) = {x|0 ≤ x}.
It is important to note that ∞ is not a real number. It is used symbolically here to indicate that this set includes all real
numbers greater than or equal to zero. Similarly, if we wanted to describe the set of all nonpositive numbers, we could write
(−∞, 0] = {x|x ≤ 0}.
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Chapter 1 | Functions and Graphs
9
Here, the notation −∞ refers to negative infinity, and it indicates that we are including all numbers less than or equal to
zero, no matter how small. The set
(−∞, ∞) = x|x is any real number
⎧
⎨
⎩
⎫
⎬
⎭
refers to the set of all real numbers.
Some functions are defined using different equations for different parts of their domain. These types of functions are known
as piecewise-defined functions. For example, suppose we want to define a function f with a domain that is the set of all
real numbers such that f (x) = 3x + 1 for x ≥ 2 and f (x) = x 2 for x < 2. We denote this function by writing
⎧3x + 1
f (x) = ⎨ 2
⎩x
x≥2
x<2
.
When evaluating this function for an input x, the equation to use depends on whether x ≥ 2 or x < 2. For example,
since 5 > 2, we use the fact that f (x) = 3x + 1 for x ≥ 2 and see that f (5) = 3(5) + 1 = 16. On the other hand, for
x = −1, we use the fact that f (x) = x 2 for x < 2 and see that f (−1) = 1.
Example 1.1
Evaluating Functions
For the function f (x) = 3x 2 + 2x − 1, evaluate
a.
f (−2)
b.
f ( 2)
c.
f (a + h)
Solution
Substitute the given value for x in the formula for f (x).
a.
f (−2) = 3(−2) 2 + 2(−2) − 1 = 12 − 4 − 1 = 7
b.
f ( 2) = 3( 2) 2 + 2 2 − 1 = 6 + 2 2 − 1 = 5 + 2 2
c.
1.1
f (a + h) = 3(a + h) 2 + 2(a + h) − 1 = 3⎛⎝a 2 + 2ah + h 2⎞⎠ + 2a + 2h − 1
= 3a 2 + 6ah + 3h 2 + 2a + 2h − 1
For f (x) = x 2 − 3x + 5, evaluate f (1) and f (a + h).
Example 1.2
Finding Domain and Range
For each of the following functions, determine the i. domain and ii. range.
10
Chapter 1 | Functions and Graphs
a.
f (x) = (x − 4) 2 + 5
b.
f (x) = 3x + 2 − 1
c.
f (x) =
3
x−2
Solution
a. Consider f (x) = (x − 4) 2 + 5.
i. Since f (x) = (x − 4) 2 + 5 is a real number for any real number x, the domain of f is the
interval (−∞, ∞).
ii. Since (x − 4) 2 ≥ 0, we know f (x) = (x − 4) 2 + 5 ≥ 5. Therefore, the range must be a subset
of y|y ≥ 5 . To show that every element in this set is in the range, we need to show that for a
⎧
⎨
⎩
⎫
⎬
⎭
given y in that set, there is a real number x such that f (x) = (x − 4) 2 + 5 = y. Solving this
equation for x, we see that we need x such that
(x − 4) 2 = y − 5.
This equation is satisfied as long as there exists a real number x such that
x − 4 = ± y − 5.
Since y ≥ 5, the square root is well-defined. We conclude that for x = 4 ± y − 5, f (x) = y,
and therefore the range is y|y ≥ 5 .
⎧
⎨
⎩
⎫
⎬
⎭
b. Consider f (x) = 3x + 2 − 1.
i. To find the domain of f , we need the expression 3x + 2 ≥ 0. Solving this inequality, we
conclude that the domain is {x|x ≥ −2/3}.
ii. To find the range of f , we note that since 3x + 2 ≥ 0, f (x) = 3x + 2 − 1 ≥ −1. Therefore,
the range of f must be a subset of the set y|y ≥ −1 . To show that every element in this set is
⎧
⎨
⎩
⎫
⎬
⎭
in the range of f , we need to show that for all y in this set, there exists a real number x in the
domain such that f (x) = y. Let y ≥ −1. Then, f (x) = y if and only if
3x + 2 − 1 = y.
Solving this equation for x, we see that x must solve the equation
3x + 2 = y + 1.
Since y ≥ −1,
such an x could exist. Squaring both sides of this equation, we have
2
3x + 2 = (y + 1) .
Therefore, we need
3x = (y + 1) 2 − 2,
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Chapter 1 | Functions and Graphs
11
which implies
x = 1 ⎛⎝y + 1⎞⎠ 2 − 2 .
3
3
We just need to verify that x is in the domain of f . Since the domain of f consists of all real
numbers greater than or equal to −2/3, and
1 ⎛⎝y + 1⎞⎠ 2 − 2 ≥ − 2 ,
3
3
3
there does exist an x in the domain of f . We conclude that the range of f is y|y ≥ −1 .
⎧
⎨
⎩
⎫
⎬
⎭
c. Consider f (x) = 3/(x − 2).
i. Since 3/(x − 2) is defined when the denominator is nonzero, the domain is {x|x ≠ 2}.
ii. To find the range of f , we need to find the values of y such that there exists a real number x
in the domain with the property that
3 = y.
x−2
Solving this equation for x, we find that
x = 3y + 2.
Therefore, as long as y ≠ 0, there exists a real number x in the domain such that f (x) = y.
Thus, the range is y|y ≠ 0 .
⎧
⎨
⎩
1.2
⎫
⎬
⎭
Find the domain and range for f (x) = 4 − 2x + 5.
Representing Functions
Typically, a function is represented using one or more of the following tools:
• A table
• A graph
• A formula
We can identify a function in each form, but we can also use them together. For instance, we can plot on a graph the values
from a table or create a table from a formula.
Tables
Functions described using a table of values arise frequently in real-world applications. Consider the following simple
example. We can describe temperature on a given day as a function of time of day. Suppose we record the temperature every
hour for a 24-hour period starting at midnight. We let our input variable x be the time after midnight, measured in hours,
and the output variable y be the temperature x hours after midnight, measured in degrees Fahrenheit. We record our data
in Table 1.1.
12
Chapter 1 | Functions and Graphs
Hours after Midnight
Temperature (°F)
Hours after Midnight
Temperature (°F)
0
58
12
84
1
54
13
85
2
53
14
85
3
52
15
83
4
52
16
82
5
55
17
80
6
60
18
77
7
64
19
74
8
72
20
69
9
75
21
65
10
78
22
60
11
80
23
58
Table 1.1 Temperature as a Function of Time of Day
We can see from the table that temperature is a function of time, and the temperature decreases, then increases, and then
decreases again. However, we cannot get a clear picture of the behavior of the function without graphing it.
Graphs
Given a function f described by a table, we can provide a visual picture of the function in the form of a graph. Graphing
the temperatures listed in Table 1.1 can give us a better idea of their fluctuation throughout the day. Figure 1.6 shows the
plot of the temperature function.
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Chapter 1 | Functions and Graphs
13
Figure 1.6 The graph of the data from Table 1.1 shows
temperature as a function of time.
From the points plotted on the graph in Figure 1.6, we can visualize the general shape of the graph. It is often useful
to connect the dots in the graph, which represent the data from the table. In this example, although we cannot make any
definitive conclusion regarding what the temperature was at any time for which the temperature was not recorded, given
the number of data points collected and the pattern in these points, it is reasonable to suspect that the temperatures at other
times followed a similar pattern, as we can see in Figure 1.7.
Figure 1.7 Connecting the dots in Figure 1.6 shows the
general pattern of the data.
Algebraic Formulas
Sometimes we are not given the values of a function in table form, rather we are given the values in an explicit formula.
Formulas arise in many applications. For example, the area of a circle of radius r is given by the formula A(r) = πr 2.
When an object is thrown upward from the ground with an initial velocity v 0 ft/s, its height above the ground from the
time it is thrown until it hits the ground is given by the formula s(t) = −16t 2 + v 0 t. When P dollars are invested in an
account at an annual interest rate r compounded continuously, the amount of money after t years is given by the formula
A(t) = Pe rt. Algebraic formulas are important tools to calculate function values. Often we also represent these functions
visually in graph form.
14
Chapter 1 | Functions and Graphs
Given an algebraic formula for a function f , the graph of f is the set of points ⎛⎝x, f (x)⎞⎠, where x is in the domain of
f and f (x) is in the range. To graph a function given by a formula, it is helpful to begin by using the formula to create
a table of inputs and outputs. If the domain of f consists of an infinite number of values, we cannot list all of them, but
because listing some of the inputs and outputs can be very useful, it is often a good way to begin.
When creating a table of inputs and outputs, we typically check to determine whether zero is an output. Those values of
x where f (x) = 0 are called the zeros of a function. For example, the zeros of f (x) = x 2 − 4 are x = ± 2. The zeros
determine where the graph of f intersects the x -axis, which gives us more information about the shape of the graph of
the function. The graph of a function may never intersect the x-axis, or it may intersect multiple (or even infinitely many)
times.
Another point of interest is the y -intercept, if it exists. The y -intercept is given by ⎛⎝0, f (0)⎞⎠.
Since a function has exactly one output for each input, the graph of a function can have, at most, one y -intercept. If x = 0
is in the domain of a function f , then f has exactly one y -intercept. If x = 0 is not in the domain of f , then f has
no y -intercept. Similarly, for any real number c, if c is in the domain of f , there is exactly one output f (c), and the
line x = c intersects the graph of f exactly once. On the other hand, if c is not in the domain of f , f (c) is not defined
and the line x = c does not intersect the graph of f . This property is summarized in the vertical line test.
Rule: Vertical Line Test
Given a function f , every vertical line that may be drawn intersects the graph of f no more than once. If any vertical
line intersects a set of points more than once, the set of points does not represent a function.
We can use this test to determine whether a set of plotted points represents the graph of a function (Figure 1.8).
Figure 1.8 (a) The set of plotted points represents the graph of
a function because every vertical line intersects the set of points,
at most, once. (b) The set of plotted points does not represent the
graph of a function because some vertical lines intersect the set
of points more than once.
Example 1.3
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Chapter 1 | Functions and Graphs
15
Finding Zeros and y -Intercepts of a Function
Consider the function f (x) = −4x + 2.
a. Find all zeros of f .
b. Find the y -intercept (if any).
c. Sketch a graph of f .
Solution
a. To find the zeros, solve f (x) = −4x + 2 = 0. We discover that f has one zero at x = 1/2.
b. The y -intercept is given by ⎛⎝0, f (0)⎞⎠ = (0, 2).
c. Given that f is a linear function of the form f (x) = mx + b that passes through the points (1/2, 0) and
(0, 2), we can sketch the graph of f (Figure 1.9).
Figure 1.9 The function f (x) = −4x + 2 is a line with
x -intercept (1/2, 0) and y -intercept (0, 2).
Example 1.4
Using Zeros and y -Intercepts to Sketch a Graph
Consider the function f (x) = x + 3 + 1.
a. Find all zeros of f .
b. Find the y -intercept (if any).
c. Sketch a graph of f .
Solution
a. To find the zeros, solve
x + 3 + 1 = 0. This equation implies x + 3 = −1. Since x + 3 ≥ 0 for all
16
Chapter 1 | Functions and Graphs
x, this equation has no solutions, and therefore f has no zeros.
b. The y -intercept is given by ⎛⎝0, f (0)⎞⎠ = (0, 3 + 1).
c. To graph this function, we make a table of values. Since we need x + 3 ≥ 0, we need to choose values
of x ≥ −3. We choose values that make the square-root function easy to evaluate.
x
−3
−2
1
f (x)
1
2
3
Table 1.2
Making use of the table and knowing that, since the function is a square root, the graph of f should be similar to
the graph of y = x, we sketch the graph (Figure 1.10).
Figure 1.10 The graph of f (x) = x + 3 + 1 has a
y -intercept but no x -intercepts.
1.3
Find the zeros of f (x) = x 3 − 5x 2 + 6x.
Example 1.5
Finding the Height of a Free-Falling Object
If a ball is dropped from a height of 100 ft, its height s at time t is given by the function s(t) = −16t 2 + 100,
where s is measured in feet and t is measured in seconds. The domain is restricted to the interval [0, c], where
t = 0 is the time when the ball is dropped and t = c is the time when the ball hits the ground.
a. Create a table showing the height s(t) when t = 0, 0.5, 1, 1.5, 2, and 2.5. Using the data from the
table, determine the domain for this function. That is, find the time c when the ball hits the ground.
b. Sketch a graph of s.
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Chapter 1 | Functions and Graphs
17
Solution
a.
t
0
0.5
1
1.5
2
2.5
s(t)
100
96
84
64
36
0
Table 1.3
Height s as a Function of Time t
Since the ball hits the ground when t = 2.5, the domain of this function is the interval [0, 2.5].
b.
Note that for this function and the function f (x) = −4x + 2 graphed in Figure 1.9, the values of f (x) are getting
smaller as x is getting larger. A function with this property is said to be decreasing. On the other hand, for the function
f (x) = x + 3 + 1 graphed in Figure 1.10, the values of f (x) are getting larger as the values of x are getting larger.
A function with this property is said to be increasing. It is important to note, however, that a function can be increasing on
some interval or intervals and decreasing over a different interval or intervals. For example, using our temperature function
in Figure 1.6, we can see that the function is decreasing on the interval (0, 4), increasing on the interval (4, 14), and
then decreasing on the interval (14, 23). We make the idea of a function increasing or decreasing over a particular interval
more precise in the next definition.
Definition
We say that a function f is increasing on the interval I if for all x 1, x 2 ∈ I,
f (x 1) ≤ f (x 2) when x 1 < x 2.
We say f is strictly increasing on the interval I if for all x 1, x 2 ∈ I,
18
Chapter 1 | Functions and Graphs
f (x 1) < f (x 2) when x 1 < x 2.
We say that a function f is decreasing on the interval I if for all x 1, x 2 ∈ I,
f (x 1) ≥ f (x 2) if x 1 < x 2.
We say that a function f is strictly decreasing on the interval I if for all x 1, x 2 ∈ I,
f (x 1) > f (x 2) if x 1 < x 2.
For example, the function f (x) = 3x is increasing on the interval (−∞, ∞) because 3x 1 < 3x 2 whenever x 1 < x 2.
On the other hand, the function f (x) = −x 3 is decreasing on the interval (−∞, ∞) because −x 13 > − x 23 whenever
x 1 < x 2 (Figure 1.11).
Figure 1.11 (a) The function f (x) = 3x is increasing on the interval (−∞, ∞). (b) The
function f (x) = −x 3 is decreasing on the interval (−∞, ∞).
Combining Functions
Now that we have reviewed the basic characteristics of functions, we can see what happens to these properties when we
combine functions in different ways, using basic mathematical operations to create new functions. For example, if the cost
for a company to manufacture x items is described by the function C(x) and the revenue created by the sale of x items is
described by the function R(x), then the profit on the manufacture and sale of x items is defined as P(x) = R(x) − C(x).
Using the difference between two functions, we created a new function.
Alternatively, we can create a new function by composing two functions. For example, given the functions f (x) = x 2 and
g(x) = 3x + 1, the composite function f ∘ g is defined such that
⎛
⎝
f ∘ g⎞⎠(x) = f ⎛⎝g(x)⎞⎠ = ⎛⎝g(x)⎞⎠ 2 = (3x + 1) 2.
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Chapter 1 | Functions and Graphs
19
The composite function g ∘ f is defined such that
g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠ = 3 f (x) + 1 = 3x 2 + 1.
⎛
⎝
Note that these two new functions are different from each other.
Combining Functions with Mathematical Operators
To combine functions using mathematical operators, we simply write the functions with the operator and simplify. Given
two functions f and g, we can define four new functions:
⎛
⎝
f + g⎞⎠(x) = f (x) + g(x)
⎛
⎞
⎝ f − g⎠(x) = f (x) − g(x)
⎛
⎞
⎝ f · g⎠(x) = f (x)g(x)
f (x)
⎛f ⎞
⎝ g ⎠(x) = g(x) for g(x) ≠ 0
Sum
Diffe ence
Product
Quotient
Example 1.6
Combining Functions Using Mathematical Operations
Given the functions f (x) = 2x − 3 and g(x) = x 2 − 1, find each of the following functions and state its
domain.
a.
( f + g)(x)
b.
( f − g)(x)
c.
( f · g)(x)
d.
⎛f ⎞
⎝ g ⎠(x)
Solution
a.
⎛
⎝
f + g⎞⎠(x) = (2x − 3) + (x 2 − 1) = x 2 + 2x − 4. The domain of this function is the interval (−∞, ∞).
b.
⎛
⎝
f − g⎞⎠(x) = (2x − 3) − (x 2 − 1) = −x 2 + 2x − 2. The domain of this function is the interval
(−∞, ∞).
c.
⎛
⎝
f · g⎞⎠(x) = (2x − 3)(x 2 − 1) = 2x 3 − 3x 2 − 2x + 3. The domain of this function is the interval
(−∞, ∞).
d.
1.4
⎛f ⎞
2x − 3
⎝ g ⎠(x) = x 2 − 1 . The domain of this function is {x|x ≠ ±1}.
For f (x) = x 2 + 3 and g(x) = 2x − 5, find ⎛⎝ f /g⎞⎠(x) and state its domain.
Function Composition
When we compose functions, we take a function of a function. For example, suppose the temperature T on a given day is
described as a function of time t (measured in hours after midnight) as in Table 1.1. Suppose the cost C, to heat or cool
a building for 1 hour, can be described as a function of the temperature T. Combining these two functions, we can describe
20
Chapter 1 | Functions and Graphs
the cost of heating or cooling a building as a function of time by evaluating C⎛⎝T(t)⎞⎠. We have defined a new function,
denoted C ∘ T, which is defined such that (C ∘ T)(t) = C(T(t)) for all t in the domain of T. This new function is called
a composite function. We note that since cost is a function of temperature and temperature is a function of time, it makes
sense to define this new function (C ∘ T)(t). It does not make sense to consider (T ∘ C)(t), because temperature is not a
function of cost.
Definition
Consider the function f with domain A and range B, and the function g with domain D and range E. If B is a
subset of D, then the composite function (g ∘ f )(x) is the function with domain A such that
g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠.
⎛
⎝
(1.1)
A composite function g ∘ f can be viewed in two steps. First, the function f maps each input x in the domain of f to
its output f (x) in the range of f . Second, since the range of f is a subset of the domain of g, the output f (x) is an
element in the domain of g, and therefore it is mapped to an output g⎛⎝ f (x)⎞⎠ in the range of g. In Figure 1.12, we see a
visual image of a composite function.
Figure 1.12 For the composite function g ∘ f , we have
⎛
⎝
g ∘ f ⎞⎠(1) = 4, ⎛⎝g ∘ f ⎞⎠(2) = 5, and ⎛⎝g ∘ f ⎞⎠(3) = 4.
Example 1.7
Compositions of Functions Defined by Formulas
Consider the functions f (x) = x 2 + 1 and g(x) = 1/x.
a. Find (g ∘ f )(x) and state its domain and range.
b. Evaluate (g ∘ f )(4), (g ∘ f )(−1/2).
c. Find ( f ∘ g)(x) and state its domain and range.
d. Evaluate ( f ∘ g)(4), ( f ∘ g)(−1/2).
Solution
a. We can find the formula for (g ∘ f )(x) in two different ways. We could write
(g ∘ f )(x) = g( f (x)) = g(x 2 + 1) =
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1 .
x2 + 1
Chapter 1 | Functions and Graphs
21
Alternatively, we could write
(g ∘ f )(x) = g⎛⎝ f (x)⎞⎠ = 1 = 2 1 .
f (x) x + 1
Since x 2 + 1 ≠ 0 for all real numbers x, the domain of (g ∘ f )(x) is the set of all real numbers. Since
0 < 1/(x 2 + 1) ≤ 1, the range is, at most, the interval (0, 1]. To show that the range is this entire
interval, we let y = 1/(x 2 + 1) and solve this equation for x to show that for all y in the interval
(0, 1], there exists a real number x such that y = 1/(x 2 + 1). Solving this equation for x, we see
that x 2 + 1 = 1/y, which implies that
x = ± 1y − 1.
If y is in the interval (0, 1], the expression under the radical is nonnegative, and therefore there exists
a real number x such that 1/(x 2 + 1) = y. We conclude that the range of g ∘ f is the interval (0, 1].
b.
(g ∘ f )(4) = g( f (4)) = g(4 2 + 1) = g(17) = 1
17
⎛
⎞
⎛
⎞
⎞⎞
⎛
⎞
⎛ ⎞
⎛ ⎛
(g ∘ f )⎝− 1 ⎠ = g⎝ f ⎝− 1 ⎠⎠ = g⎜⎝− 1 ⎠ + 1⎟ = g⎝5 ⎠ = 4
2
4
2
5
⎝ 2
⎠
2
c. We can find a formula for ( f ∘ g)(x) in two ways. First, we could write
2
( f ∘ g)(x) = f (g(x)) = f ⎛⎝1x ⎞⎠ = ⎛⎝1x ⎞⎠ + 1.
Alternatively, we could write
2
( f ∘ g)(x) = f (g(x)) = (g(x)) 2 + 1 = ⎛⎝1x ⎞⎠ + 1.
The domain of f ∘ g is the set of all real numbers x such that x ≠ 0. To find the range of f , we need
to find all values y for which there exists a real number x ≠ 0 such that
⎛1 ⎞
⎝x ⎠
2
+ 1 = y.
Solving this equation for x, we see that we need x to satisfy
⎛1 ⎞
⎝x ⎠
2
= y − 1,
which simplifies to
1 = ± y − 1.
x
Finally, we obtain
22
Chapter 1 | Functions and Graphs
x= ±
1 .
y−1
Since 1/ y − 1 is a real number if and only if y > 1, the range of f is the set y|y ≥ 1 .
⎧
⎨
⎩
d.
⎫
⎬
⎭
⎛ ⎞ ⎛ ⎞
( f ∘ g)(4) = f (g(4)) = f ⎝1 ⎠ = ⎝1 ⎠ + 1 = 17
4
4
16
⎞
⎞⎞
⎛
⎛ ⎛
( f ∘ g)⎝− 1 ⎠ = f ⎝g⎝− 1 ⎠⎠ = f (−2) = (−2) 2 + 1 = 5
2
2
2
In Example 1.7, we can see that ⎛⎝ f ∘ g⎞⎠(x) ≠ ⎛⎝g ∘ f ⎞⎠(x). This tells us, in general terms, that the order in which we compose
functions matters.
1.5
Let f (x) = 2 − 5x. Let g(x) = x. Find ⎛⎝ f ∘ g⎞⎠(x).
Example 1.8
Composition of Functions Defined by Tables
Consider the functions f and g described by Table 1.4 and Table 1.5.
x
−3
−2
−1
0
1
2
3
4
f (x)
0
4
2
4
−2
0
−2
4
Table 1.4
x
−4
−2
0
2
4
g(x)
1
0
3
0
5
Table 1.5
a. Evaluate (g ∘ f )(3), ⎛⎝g ∘ f ⎞⎠(0).
b. State the domain and range of ⎛⎝g ∘ f ⎞⎠(x).
c. Evaluate ( f ∘ f )(3), ⎛⎝ f ∘ f ⎞⎠(1).
d. State the domain and range of ⎛⎝ f ∘ f ⎞⎠(x).
Solution
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Chapter 1 | Functions and Graphs
a.
23
g ∘ f ⎞⎠(3) = g⎛⎝ f (3)⎞⎠ = g(−2) = 0
⎛
⎝
(g ∘ f )(0) = g(4) = 5
b. The domain of g ∘ f is the set {−3, −2, −1, 0, 1, 2, 3, 4}. Since the range of f is the set
{−2, 0, 2, 4}, the range of g ∘ f is the set {0, 3, 5}.
c.
⎛
⎝
f ∘ f ⎞⎠(3) = f ⎛⎝ f (3)⎞⎠ = f (−2) = 4
( f ∘ f )(1) = f ( f (1)) = f (−2) = 4
d. The domain of f ∘ f is the set {−3, −2, −1, 0, 1, 2, 3, 4}. Since the range of f is the set
{−2, 0, 2, 4}, the range of f ∘ f is the set {0, 4}.
Example 1.9
Application Involving a Composite Function
A store is advertising a sale of 20% off all merchandise. Caroline has a coupon that entitles her to an additional
15% off any item, including sale merchandise. If Caroline decides to purchase an item with an original price of
x dollars, how much will she end up paying if she applies her coupon to the sale price? Solve this problem by
using a composite function.
Solution
Since the sale price is 20% off the original price, if an item is x dollars, its sale price is given by f (x) = 0.80x.
Since the coupon entitles an individual to 15% off the price of any item, if an item is y dollars, the price, after
applying the coupon, is given by g(y) = 0.85y. Therefore, if the price is originally x dollars, its sale price will
be f (x) = 0.80x and then its final price after the coupon will be g( f (x)) = 0.85(0.80x) = 0.68x.
1.6 If items are on sale for 10% off their original price, and a customer has a coupon for an additional 30%
off, what will be the final price for an item that is originally x dollars, after applying the coupon to the sale
price?
Symmetry of Functions
The graphs of certain functions have symmetry properties that help us understand the function and the shape of its graph.
For example, consider the function f (x) = x 4 − 2x 2 − 3 shown in Figure 1.13(a). If we take the part of the curve that
lies to the right of the y-axis and flip it over the y-axis, it lays exactly on top of the curve to the left of the y-axis. In this
case, we say the function has symmetry about the y-axis. On the other hand, consider the function f (x) = x 3 − 4x shown
in Figure 1.13(b). If we take the graph and rotate it 180° about the origin, the new graph will look exactly the same. In
this case, we say the function has symmetry about the origin.
24
Chapter 1 | Functions and Graphs
Figure 1.13 (a) A graph that is symmetric about the y -axis. (b) A graph that is symmetric
about the origin.
If we are given the graph of a function, it is easy to see whether the graph has one of these symmetry properties. But without
a graph, how can we determine algebraically whether a function f has symmetry? Looking at Figure 1.14 again, we see
that since f is symmetric about the y -axis, if the point (x, y) is on the graph, the point (−x, y) is on the graph. In other
words, f (−x) = f (x). If a function f has this property, we say f is an even function, which has symmetry about the
y-axis. For example, f (x) = x 2 is even because
f (−x) = (−x) 2 = x 2 = f (x).
In contrast, looking at Figure 1.14 again, if a function f is symmetric about the origin, then whenever the point (x, y) is
on the graph, the point (−x, −y) is also on the graph. In other words, f (−x) = − f (x). If f has this property, we say f
is an odd function, which has symmetry about the origin. For example, f (x) = x 3 is odd because
f (−x) = (−x) 3 = −x 3 = − f (x).
Definition
If f (x) = f (−x) for all x in the domain of f , then f is an even function. An even function is symmetric about the
y-axis.
If f (−x) = − f (x) for all x in the domain of f , then f is an odd function. An odd function is symmetric about the
origin.
Example 1.10
Even and Odd Functions
Determine whether each of the following functions is even, odd, or neither.
a.
f (x) = −5x 4 + 7x 2 − 2
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Chapter 1 | Functions and Graphs
25
b.
f (x) = 2x 5 − 4x + 5
c.
f (x) =
3x
x2 + 1
Solution
To determine whether a function is even or odd, we evaluate f (−x) and compare it to f(x) and − f (x).
a.
f (−x) = −5(−x) 4 + 7(−x) 2 − 2 = −5x 4 + 7x 2 − 2 = f (x). Therefore, f is even.
b.
f (−x) = 2(−x) 5 − 4(−x) + 5 = −2x 5 + 4x + 5. Now,
f (−x) ≠ f (x). Furthermore, noting that
− f (x) = −2x 5 + 4x − 5, we see that f (−x) ≠ − f (x). Therefore, f is neither even nor odd.
c.
1.7
f (−x) = 3(−x)/((−x) 2 + 1} = −3x/(x 2 + 1) = −[3x/(x 2 + 1)] = − f (x). Therefore, f is odd.
Determine whether f (x) = 4x 3 − 5x is even, odd, or neither.
One symmetric function that arises frequently is the absolute value function, written as |x|. The absolute value function is
defined as
⎧−x, x < 0
f (x) = ⎨
.
⎩ x, x ≥ 0
(1.2)
Some students describe this function by stating that it “makes everything positive.” By the definition of the absolute value
function, we see that if x < 0, then |x| = −x > 0, and if x > 0, then |x| = x > 0. However, for x = 0, |x| = 0.
Therefore, it is more accurate to say that for all nonzero inputs, the output is positive, but if x = 0, the output |x| = 0. We
conclude that the range of the absolute value function is y|y ≥ 0 . In Figure 1.14, we see that the absolute value function
⎧
⎨
⎩
⎫
⎬
⎭
is symmetric about the y-axis and is therefore an even function.
Figure 1.14 The graph of f (x) = |x| is symmetric about the
y -axis.
26
Chapter 1 | Functions and Graphs
Example 1.11
Working with the Absolute Value Function
Find the domain and range of the function f (x) = 2|x − 3| + 4.
Solution
Since the absolute value function is defined for all real numbers, the domain of this function is (−∞, ∞). Since
| x − 3| ≥ 0
for all x, the function f (x) = 2|x − 3| + 4 ≥ 4. Therefore, the range is, at most, the set y|y ≥ 4 .
⎧
⎨
⎩
⎫
⎬
⎭
To see that the range is, in fact, this whole set, we need to show that for y ≥ 4 there exists a real number x such
that
2|x − 3| + 4 = y.
A real number x satisfies this equation as long as
|x − 3| = 1 (y − 4).
2
Since y ≥ 4, we know y − 4 ≥ 0, and thus the right-hand side of the equation is nonnegative, so it is possible
that there is a solution. Furthermore,
⎧−(x − 3) if x < 3
.
|x − 3| = ⎨
⎩x−3
if x ≥ 3
Therefore, we see there are two solutions:
x = ± 1 (y − 4) + 3.
2
The range of this function is y|y ≥ 4 .
⎧
⎨
⎩
1.8
⎫
⎬
⎭
For the function f (x) = |x + 2| − 4, find the domain and range.
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Chapter 1 | Functions and Graphs
27
1.1 EXERCISES
For the following exercises, (a) determine the domain and
the range of each relation, and (b) state whether the relation
is a function.
4.
1.
x
y
x
y
−3
9
1
1
−2
4
2
4
−1
1
3
9
0
0
x
y
x
y
1
1
5
1
2
1
6
1
3
1
7
1
4
1
x
y
x
y
3
3
15
1
5
2
21
2
8
1
33
3
10
0
x
y
x
y
−7
11
1
−2
−2
5
3
4
−2
1
6
11
0
−1
5.
2.
x
y
x
y
−3
−2
1
1
−2
−8
2
8
−1
−1
3
−2
0
0
6.
3.
x
y
x
y
1
−3
1
1
2
−2
2
2
3
−1
3
3
0
0
For the following exercises, find the values for each
function, if they exist, then simplify.
a. f (0) b. f (1) c. f (3) d. f (−x) e. f (a) f. f (a + h)
7.
f (x) = 5x − 2
28
Chapter 1 | Functions and Graphs
8.
f (x) = 4x 2 − 3x + 1
9.
f (x) = 2x
23.
f (x) = 3x − 6
x
y
x
y
−3
−15
1
−3
10.
f (x) = |x − 7| + 8
11.
f (x) = 6x + 5
−2
−12
2
0
12.
f (x) = x − 2
3x + 7
−1
−9
3
3
13.
f (x) = 9
0
−6
For the following exercises, find the domain, range, and all
zeros/intercepts, if any, of the functions.
14.
f (x) =
24.
x
x 2 − 16
f (x) = 1 x + 1
2
x
y
x
y
−3
−1
2
1
3
2
−2
0
2
2
−1
1
2
3
5
2
0
1
15. g(x) = 8x − 1
16. h(x) = 2 3
x +4
17.
f (x) = −1 + x + 2
18.
f (x) =
19. g(x) =
20.
1
x−9
3
x−4
f (x) = 4|x + 5|
21. g(x) =
25.
7
x−5
For the following exercises, set up a table to sketch the
graph of each function using the following values:
x = −3, −2, −1, 0, 1, 2, 3.
22.
f (x) = x 2 + 1
x
y
x
y
−3
10
1
2
−2
5
2
5
−1
2
3
10
0
1
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f (x) = 2|x|
x
y
x
y
−3
6
1
2
−2
4
2
4
−1
2
3
6
0
0
Chapter 1 | Functions and Graphs
29
f (x) = −x 2
26.
28.
x
y
x
y
−3
−9
1
−1
−2
−4
2
−4
−1
−1
3
−9
0
0
x
y
x
y
−3
−27
1
1
−2
−8
2
8
−1
−1
3
27
0
0
f (x) = x 3
27.
29.
For the following exercises, use the vertical line test to
determine whether each of the given graphs represents a
function. Assume that a graph continues at both ends if
it extends beyond the given grid. If the graph represents a
function, then determine the following for each graph:
a. Domain and range
b.
x -intercept, if any (estimate where necessary)
c.
y -Intercept, if any (estimate where necessary)
d. The intervals for which the function is increasing
e. The intervals for which the function is decreasing
f. The intervals for which the function is constant
g. Symmetry about any axis and/or the origin
h. Whether the function is even, odd, or neither
30.
30
31.
Chapter 1 | Functions and Graphs
34.
32.
35.
33.
For the following exercises, for each pair of functions, find
a. f + g b. f − g c. f · g d. f /g. Determine the domain
of each of these new functions.
36.
f (x) = 3x + 4, g(x) = x − 2
37.
f (x) = x − 8, g(x) = 5x 2
38.
f (x) = 3x 2 + 4x + 1, g(x) = x + 1
39.
f (x) = 9 − x 2, g(x) = x 2 − 2x − 3
40.
f (x) = x, g(x) = x − 2
41.
f (x) = 6 + 1x , g(x) = 1x
For the following exercises, for each pair of functions, find
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Chapter 1 | Functions and Graphs
a. ⎛⎝ f ∘ g⎞⎠(x) and b. ⎛⎝g ∘ f ⎞⎠(x) Simplify the results. Find the
domain of each of the results.
31
49. The table below lists the NBA championship winners
for the years 2001 to 2012.
Year
Winner
2001
LA Lakers
f (x) = 2x + 4, g(x) = x 2 − 2
2002
LA Lakers
45.
f (x) = x 2 + 7, g(x) = x 2 − 3
2003
San Antonio Spurs
46.
f (x) = x, g(x) = x + 9
2004
Detroit Pistons
47.
f (x) =
3 , g(x) = 2
x
2x + 1
2005
San Antonio Spurs
48.
f (x) = |x + 1|, g(x) = x 2 + x − 4
2006
Miami Heat
2007
San Antonio Spurs
2008
Boston Celtics
2009
LA Lakers
2010
LA Lakers
2011
Dallas Mavericks
2012
Miami Heat
42.
f (x) = 3x, g(x) = x + 5
43.
f (x) = x + 4, g(x) = 4x − 1
44.
a. Consider the relation in which the domain values
are the years 2001 to 2012 and the range is the
corresponding winner. Is this relation a function?
Explain why or why not.
b. Consider the relation where the domain values are
the winners and the range is the corresponding
years. Is this relation a function? Explain why or
why not.
50. [T] The area A of a square depends on the length of
the side s.
a. Write a function A(s) for the area of a square.
b. Find and interpret A(6.5).
c. Find the exact and the two-significant-digit
approximation to the length of the sides of a square
with area 56 square units.
32
Chapter 1 | Functions and Graphs
51. [T] The volume of a cube depends on the length of the
sides s.
a. Write a function V(s) for the area of a square.
b. Find and interpret V(11.8).
52. [T] A rental car company rents cars for a flat fee of
$20 and an hourly charge of $10.25. Therefore, the total
cost C to rent a car is a function of the hours t the car is
rented plus the flat fee.
a. Write the formula for the function that models this
situation.
b. Find the total cost to rent a car for 2 days and 7
hours.
c. Determine how long the car was rented if the bill is
$432.73.
53. [T] A vehicle has a 20-gal tank and gets 15 mpg.
The number of miles N that can be driven depends on the
amount of gas x in the tank.
a. Write a formula that models this situation.
b. Determine the number of miles the vehicle can
travel on (i) a full tank of gas and (ii) 3/4 of a tank
of gas.
c. Determine the domain and range of the function.
d. Determine how many times the driver had to stop
for gas if she has driven a total of 578 mi.
54. [T] The volume V of a sphere depends on the length of
its radius as V = (4/3)πr 3. Because Earth is not a perfect
sphere, we can use the mean radius when measuring from
the center to its surface. The mean radius is the average
distance from the physical center to the surface, based on
a large number of samples. Find the volume of Earth with
mean radius 6.371 × 10 6 m.
55. [T] A certain bacterium grows in culture in a circular
region. The radius of the circle, measured in centimeters,
⎡
⎤
is given by r(t) = 6 − ⎣5/⎛⎝t 2 + 1⎞⎠⎦, where t is time
measured in hours since a circle of a 1-cm radius of the
bacterium was put into the culture.
a. Express the area of the bacteria as a function of
time.
b. Find the exact and approximate area of the bacterial
culture in 3 hours.
c. Express the circumference of the bacteria as a
function of time.
d. Find the exact and approximate circumference of
the bacteria in 3 hours.
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56. [T] An American tourist visits Paris and must convert
U.S. dollars to Euros, which can be done using the function
E(x) = 0.79x, where x is the number of U.S. dollars and
E(x) is the equivalent number of Euros. Since conversion
rates fluctuate, when the tourist returns to the United States
2 weeks later, the conversion from Euros to U.S. dollars
is D(x) = 1.245x, where x is the number of Euros and
D(x) is the equivalent number of U.S. dollars.
a. Find the composite function that converts directly
from U.S. dollars to U.S. dollars via Euros. Did this
tourist lose value in the conversion process?
b. Use (a) to determine how many U.S. dollars the
tourist would get back at the end of her trip if she
converted an extra $200 when she arrived in Paris.
57. [T] The manager at a skateboard shop pays his
workers a monthly salary S of $750 plus a commission of
$8.50 for each skateboard they sell.
a. Write a function y = S(x) that models a worker’s
monthly salary based on the number of skateboards
x he or she sells.
b. Find the approximate monthly salary when a
worker sells 25, 40, or 55 skateboards.
c. Use the INTERSECT feature on a graphing
calculator to determine the number of skateboards
that must be sold for a worker to earn a monthly
income of $1400. (Hint: Find the intersection of the
function and the line y = 1400.)
Chapter 1 | Functions and Graphs
58. [T] Use a graphing calculator to graph the half-circle
y = 25 − (x − 4) 2. Then, use the INTERCEPT feature
to find the value of both the x - and y -intercepts.
33
34
Chapter 1 | Functions and Graphs
1.2 | Basic Classes of Functions
Learning Objectives
1.2.1 Calculate the slope of a linear function and interpret its meaning.
1.2.2 Recognize the degree of a polynomial.
1.2.3 Find the roots of a quadratic polynomial.
1.2.4 Describe the graphs of basic odd and even polynomial functions.
1.2.5 Identify a rational function.
1.2.6 Describe the graphs of power and root functions.
1.2.7 Explain the difference between algebraic and transcendental functions.
1.2.8 Graph a piecewise-defined function.
1.2.9 Sketch the graph of a function that has been shifted, stretched, or reflected from its initial
graph position.
We have studied the general characteristics of functions, so now let’s examine some specific classes of functions. We
begin by reviewing the basic properties of linear and quadratic functions, and then generalize to include higher-degree
polynomials. By combining root functions with polynomials, we can define general algebraic functions and distinguish
them from the transcendental functions we examine later in this chapter. We finish the section with examples of piecewisedefined functions and take a look at how to sketch the graph of a function that has been shifted, stretched, or reflected from
its initial form.
Linear Functions and Slope
The easiest type of function to consider is a linear function. Linear functions have the form f (x) = ax + b, where a and
b are constants. In Figure 1.15, we see examples of linear functions when a is positive, negative, and zero. Note that if
a > 0, the graph of the line rises as x increases. In other words, f (x) = ax + b is increasing on (−∞, ∞). If a < 0,
the graph of the line falls as x increases. In this case, f (x) = ax + b is decreasing on (−∞, ∞). If a = 0, the line is
horizontal.
Figure 1.15 These linear functions are increasing or
decreasing on (∞, ∞) and one function is a horizontal line.
As suggested by Figure 1.15, the graph of any linear function is a line. One of the distinguishing features of a line is its
slope. The slope is the change in y for each unit change in x. The slope measures both the steepness and the direction of
a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line points
downward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, we
need to determine the ratio of the change in y versus the change in x. To do so, we choose any two points (x 1, y 1) and
y −y
(x 2, y 2) on the line and calculate x 2 − x 1 . In Figure 1.16, we see this ratio is independent of the points chosen.
2
1
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Chapter 1 | Functions and Graphs
35
Figure 1.16 For any linear function, the slope
(y 2 − y 1)/(x 2 − x 1) is independent of the choice of points
(x 1, y 1) and (x 2, y 2) on the line.
Definition
Consider line L passing through points (x 1, y 1) and (x 2, y 2). Let Δy = y 2 − y 1 and Δx = x 2 − x 1 denote the
changes in y and x, respectively. The slope of the line is
y −y
Δy
m = x2 − x1 =
.
Δx
2
1
(1.3)
We now examine the relationship between slope and the formula for a linear function. Consider the linear function given
by the formula f (x) = ax + b. As discussed earlier, we know the graph of a linear function is given by a line. We
can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating
(y 2 − y 1)/(x 2 − x 1) for any points (x 1, y 1) and (x 2, y 2) on the line. Evaluating the function f at x = 0, we see
that (0, b) is a point on this line. Evaluating this function at x = 1, we see that (1, a + b) is also a point on this line.
Therefore, the slope of this line is
(a + b) − b
= a.
1−0
We have shown that the coefficient a is the slope of the line. We can conclude that the formula f (x) = ax + b describes
a line with slope a. Furthermore, because this line intersects the y -axis at the point (0, b), we see that the y -intercept
for this linear function is (0, b). We conclude that the formula f (x) = ax + b tells us the slope, a, and the y -intercept,
(0, b), for this line. Since we often use the symbol m to denote the slope of a line, we can write
f (x) = mx + b
to denote the slope-intercept form of a linear function.
Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear function
passes through the point (x 1, y 1) and the slope of the line is m. Since any other point (x, f (x)) on the graph of f must
satisfy the equation
f (x) − y
m = x − x 1,
1
36
Chapter 1 | Functions and Graphs
this linear function can be expressed by writing
f (x) − y 1 = m(x − x 1).
We call this equation the point-slope equation for that linear function.
Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using the
slope-intercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot be
expressed in either of these forms. Instead, a vertical line is described by the equation x = k for some constant k. Since
neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation
ax + by = c,
where a, b are both not zero, to denote the standard form of a line.
Definition
Consider a line passing through the point (x 1, y 1) with slope m. The equation
y − y 1 = m(x − x 1)
(1.4)
is the point-slope equation for that line.
Consider a line with slope m and y -intercept (0, b). The equation
y = mx + b
(1.5)
is an equation for that line in slope-intercept form.
The standard form of a line is given by the equation
ax + by = c,
where a and b are both not zero. This form is more general because it allows for a vertical line, x = k.
Example 1.12
Finding the Slope and Equations of Lines
Consider the line passing through the points (11, −4) and (−4, 5), as shown in Figure 1.17.
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(1.6)
Chapter 1 | Functions and Graphs
37
Figure 1.17 Finding the equation of a linear function with a graph that is a line between
two given points.
a. Find the slope of the line.
b. Find an equation for this linear function in point-slope form.
c. Find an equation for this linear function in slope-intercept form.
Solution
a. The slope of the line is
y −y
5 − (−4)
m = x2 − x1 =
= − 9 = − 3.
−4 − 11
15
5
2
1
b. To find an equation for the linear function in point-slope form, use the slope m = −3/5 and choose any
point on the line. If we choose the point (11, −4), we get the equation
f (x) + 4 = − 3 (x − 11).
5
c. To find an equation for the linear function in slope-intercept form, solve the equation in part b. for f (x).
When we do this, we get the equation
f (x) = − 3 x + 13 .
5
5
1.9
Consider the line passing through points (−3, 2) and (1, 4). Find the slope of the line.
Find an equation of that line in point-slope form. Find an equation of that line in slope-intercept form.
Example 1.13
A Linear Distance Function
Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the
following questions, assuming Jessica runs at a constant pace.
38
Chapter 1 | Functions and Graphs
a. Describe the distance D (in miles) Jessica runs as a linear function of her run time t (in minutes).
b. Sketch a graph of D.
c. Interpret the meaning of the slope.
Solution
a. At time t = 0, Jessica is at her house, so D(0) = 0. At time t = 78 minutes, Jessica has finished
running 9 mi, so D(78) = 9. The slope of the linear function is
m= 9−0 = 3 .
78 − 0 26
The y -intercept is (0, 0), so the equation for this linear function is
D(t) = 3 t.
26
b. To graph D, use the fact that the graph passes through the origin and has slope m = 3/26.
c. The slope m = 3/26 ≈ 0.115 describes the distance (in miles) Jessica runs per minute, or her average
velocity.
Polynomials
A linear function is a special type of a more general class of functions: polynomials. A polynomial function is any function
that can be written in the form
f (x) = a n x n + a n − 1 x n − 1 + … + a 1 x + a 0
(1.7)
for some integer n ≥ 0 and constants a n, a n − 1 ,…,a 0, where a n ≠ 0. In the case when n = 0, we allow for a 0 = 0;
if a 0 = 0, the function f (x) = 0 is called the zero function. The value n is called the degree of the polynomial; the
constant a n is called the leading coefficient. A linear function of the form f (x) = mx + b is a polynomial of degree 1
if m ≠ 0 and degree 0 if m = 0. A polynomial of degree 0 is also called a constant function. A polynomial function
of degree 2 is called a quadratic function. In particular, a quadratic function has the form f (x) = ax 2 + bx + c, where
a ≠ 0. A polynomial function of degree 3 is called a cubic function.
Power Functions
Some polynomial functions are power functions. A power function is any function of the form f (x) = ax b, where a and
b are any real numbers. The exponent in a power function can be any real number, but here we consider the case when the
exponent is a positive integer. (We consider other cases later.) If the exponent is a positive integer, then f (x) = ax n is a
polynomial. If n is even, then f (x) = ax n is an even function because f (−x) = a(−x) n = ax n if n is even. If n is odd,
then f (x) = ax n is an odd function because f (−x) = a(−x) n = −ax n if n is odd (Figure 1.18).
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Chapter 1 | Functions and Graphs
39
Figure 1.18 (a) For any even integer n, f (x) = ax n is an even function. (b) For any odd
integer n, f (x) = ax n is an odd function.
Behavior at Infinity
To determine the behavior of a function f as the inputs approach infinity, we look at the values f (x) as the inputs,
x, become larger. For some functions, the values of f (x) approach a finite number. For example, for the function
f (x) = 2 + 1/x, the values 1/x become closer and closer to zero for all values of x as they get larger and larger. For this
function, we say “ f (x) approaches two as x goes to infinity,” and we write f (x) → 2 as x → ∞. The line y = 2 is a
horizontal asymptote for the function f (x) = 2 + 1/x because the graph of the function gets closer to the line as x gets
larger.
For other functions, the values f (x) may not approach a finite number but instead may become larger for all values of x
as they get larger. In that case, we say “ f (x) approaches infinity as x approaches infinity,” and we write f (x) → ∞ as
x → ∞. For example, for the function f (x) = 3x 2, the outputs f (x) become larger as the inputs x get larger. We can
conclude that the function f (x) = 3x 2 approaches infinity as x approaches infinity, and we write 3x 2 → ∞ as x → ∞.
The behavior as x → −∞ and the meaning of f (x) → −∞ as x → ∞ or x → −∞ can be defined similarly. We can
describe what happens to the values of f (x) as x → ∞ and as x → −∞ as the end behavior of the function.
To understand the end behavior for polynomial functions, we can focus on quadratic and cubic functions. The behavior for
higher-degree polynomials can be analyzed similarly. Consider a quadratic function f (x) = ax 2 + bx + c. If a > 0, the
values f (x) → ∞ as x → ±∞. If a < 0, the values f (x) → −∞ as x → ±∞. Since the graph of a quadratic function
is a parabola, the parabola opens upward if a > 0; the parabola opens downward if a < 0. (See Figure 1.19(a).)
Now consider a cubic function f (x) = ax 3 + bx 2 + cx + d. If a > 0, then f (x) → ∞ as x → ∞ and f (x) → −∞
as x → −∞. If a < 0, then f (x) → −∞ as x → ∞ and f (x) → ∞ as x → −∞. As we can see from both of these
graphs, the leading term of the polynomial determines the end behavior. (See Figure 1.19(b).)
40
Chapter 1 | Functions and Graphs
Figure 1.19 (a) For a quadratic function, if the leading coefficient a > 0, the parabola opens
upward. If a < 0, the parabola opens downward. (b) For a cubic function f , if the leading
coefficient a > 0, the values f (x) → ∞ as x → ∞ and the values f (x) → −∞ as
x → −∞. If the leading coefficient a < 0, the opposite is true.
Zeros of Polynomial Functions
Another characteristic of the graph of a polynomial function is where it intersects the x -axis. To determine where a function
f intersects the x -axis, we need to solve the equation f (x) = 0 for .n the case of the linear function f (x) = mx + b,
the x -intercept is given by solving the equation mx + b = 0. In this case, we see that the x -intercept is given by
(−b/m, 0). In the case of a quadratic function, finding the x -intercept(s) requires finding the zeros of a quadratic equation:
ax 2 + bx + c = 0. In some cases, it is easy to factor the polynomial ax 2 + bx + c to find the zeros. If not, we make use
of the quadratic formula.
Rule: The Quadratic Formula
Consider the quadratic equation
ax 2 + bx + c = 0,
where a ≠ 0. The solutions of this equation are given by the quadratic formula
2
x = −b ± b − 4ac .
2a
(1.8)
If the discriminant b 2 − 4ac > 0, this formula tells us there are two real numbers that satisfy the quadratic equation.
If b 2 − 4ac = 0, this formula tells us there is only one solution, and it is a real number. If b 2 − 4ac < 0, no real
numbers satisfy the quadratic equation.
In the case of higher-degree polynomials, it may be more complicated to determine where the graph intersects the x -axis.
In some instances, it is possible to find the x -intercepts by factoring the polynomial to find its zeros. In other cases, it is
impossible to calculate the exact values of the x -intercepts. However, as we see later in the text, in cases such as this, we
can use analytical tools to approximate (to a very high degree) where the x -intercepts are located. Here we focus on the
graphs of polynomials for which we can calculate their zeros explicitly.
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Chapter 1 | Functions and Graphs
41
Example 1.14
Graphing Polynomial Functions
For the following functions a. and b., i. describe the behavior of f (x) as x → ±∞, ii. find all zeros of f , and
iii. sketch a graph of f .
a.
f (x) = −2x 2 + 4x − 1
b.
f (x) = x 3 − 3x 2 − 4x
Solution
a. The function f (x) = −2x 2 + 4x − 1 is a quadratic function.
i. Because a = −2 < 0, as x → ±∞, f (x) → −∞.
ii. To find the zeros of f , use the quadratic formula. The zeros are
x=
−4 ± 4 2 − 4(−2)(−1) −4 ± 8 −4 ± 2 2 2 ± 2
=
=
=
.
−4
−4
2
2(−2)
iii. To sketch the graph of f , use the information from your previous answers and combine it with
the fact that the graph is a parabola opening downward.
b. The function f (x) = x 3 − 3x 2 − 4x is a cubic function.
i. Because a = 1 > 0, as x → ∞, f (x) → ∞. As x → −∞, f (x) → −∞.
ii. To find the zeros of f , we need to factor the polynomial. First, when we factor x out of all the
terms, we find
f (x) = x(x 2 − 3x − 4).
Then, when we factor the quadratic function x 2 − 3x − 4, we find
f (x) = x(x − 4)(x + 1).
42
Chapter 1 | Functions and Graphs
Therefore, the zeros of f are x = 0, 4, −1.
iii. Combining the results from parts i. and ii., draw a rough sketch of f .
1.10
Consider the quadratic function f (x) = 3x 2 − 6x + 2. Find the zeros of f . Does the parabola open
upward or downward?
Mathematical Models
A large variety of real-world situations can be described using mathematical models. A mathematical model is a method of
simulating real-life situations with mathematical equations. Physicists, engineers, economists, and other researchers develop
models by combining observation with quantitative data to develop equations, functions, graphs, and other mathematical
tools to describe the behavior of various systems accurately. Models are useful because they help predict future outcomes.
Examples of mathematical models include the study of population dynamics, investigations of weather patterns, and
predictions of product sales.
As an example, let’s consider a mathematical model that a company could use to describe its revenue for the sale of a
particular item. The amount of revenue R a company receives for the sale of n items sold at a price of p dollars per item
is described by the equation R = p · n. The company is interested in how the sales change as the price of the item changes.
Suppose the data in Table 1.6 show the number of units a company sells as a function of the price per item.
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Chapter 1 | Functions and Graphs
43
p
6
8
10
12
14
n
19.4
18.5
16.2
13.8
12.2
Table 1.6 Number of Units Sold n (in Thousands) as a
Function of Price per Unit p (in Dollars)
In Figure 1.20, we see the graph the number of units sold (in thousands) as a function of price (in dollars). We note from
the shape of the graph that the number of units sold is likely a linear function of price per item, and the data can be closely
approximated by the linear function n = −1.04p + 26 for 0 ≤ p ≤ 25, where n predicts the number of units sold in
thousands. Using this linear function, the revenue (in thousands of dollars) can be estimated by the quadratic function
R(p) = p · ⎛⎝−1.04p + 26⎞⎠ = −1.04p 2 + 26p
for 0 ≤ p ≤ 25. In Example 1.15, we use this quadratic function to predict the amount of revenue the company receives
depending on the price the company charges per item. Note that we cannot conclude definitively the actual number of units
sold for values of p, for which no data are collected. However, given the other data values and the graph shown, it seems
reasonable that the number of units sold (in thousands) if the price charged is p dollars may be close to the values predicted
by the linear function n = −1.04p + 26.
Figure 1.20 The data collected for the number of items sold as a function of
price is roughly linear. We use the linear function n = −1.04p + 26 to estimate
this function.
Example 1.15
44
Chapter 1 | Functions and Graphs
Maximizing Revenue
A company is interested in predicting the amount of revenue it will receive depending on the price it charges for a
particular item. Using the data from Table 1.6, the company arrives at the following quadratic function to model
revenue R as a function of price per item p:
R(p) = p · ⎛⎝−1.04p + 26⎞⎠ = −1.04p 2 + 26p
for 0 ≤ p ≤ 25.
a. Predict the revenue if the company sells the item at a price of p = $5 and p = $17.
b. Find the zeros of this function and interpret the meaning of the zeros.
c. Sketch a graph of R.
d. Use the graph to determine the value of p that maximizes revenue. Find the maximum revenue.
Solution
a. Evaluating the revenue function at p = 5 and p = 17, we can conclude that
R(5) = −1.04(5) 2 + 26(5) = 104, so revenue = $104,000;
R(17) = −1.04(17) 2 + 26(17) = 141.44, so revenue = $144,440.
b. The zeros of this function can be found by solving the equation −1.04p 2 + 26p = 0. When we factor
the quadratic expression, we get p⎛⎝−1.04p + 26⎞⎠ = 0. The solutions to this equation are given by
p = 0, 25. For these values of p, the revenue is zero. When p = $0, the revenue is zero because the
company is giving away its merchandise for free. When p = $25, the revenue is zero because the price
is too high, and no one will buy any items.
c. Knowing the fact that the function is quadratic, we also know the graph is a parabola. Since the
leading coefficient is negative, the parabola opens downward. One property of parabolas is that they are
symmetric about the axis, so since the zeros are at p = 0 and p = 25, the parabola must be symmetric
about the line halfway between them, or p = 12.5.
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Chapter 1 | Functions and Graphs
45
d. The function is a parabola with zeros at p = 0 and p = 25,
and it is symmetric about the line
p = 12.5, so the maximum revenue occurs at a price of p = $12.50 per item. At that price, the revenue
is R(p) = −1.04(12.5) 2 + 26(12.5) = $162, 500.
Algebraic Functions
By allowing for quotients and fractional powers in polynomial functions, we create a larger class of functions. An algebraic
function is one that involves addition, subtraction, multiplication, division, rational powers, and roots. Two types of
algebraic functions are rational functions and root functions.
Just as rational numbers are quotients of integers, rational functions are quotients of polynomials. In particular, a rational
function is any function of the form f (x) = p(x)/q(x), where p(x) and q(x) are polynomials. For example,
f (x) = 3x − 1
5x + 2
and
g(x) =
4
x2 + 1
are rational functions. A root function is a power function of the form f (x) = x 1/n, where n is a positive integer greater
than one. For example, f (x) = x 1/2 = x is the square-root function and g(x) = x 1/3 = 3 x is the cube-root function. By
allowing for compositions of root functions and rational functions, we can create other algebraic functions. For example,
f (x) = 4 − x 2 is an algebraic function.
46
Chapter 1 | Functions and Graphs
Example 1.16
Finding Domain and Range for Algebraic Functions
For each of the following functions, find the domain and range.
a.
f (x) = 3x − 1
5x + 2
b.
f (x) = 4 − x 2
Solution
a. It is not possible to divide by zero, so the domain is the set of real numbers x such that x ≠ −2/5. To
find the range, we need to find the values y for which there exists a real number x such that
y = 3x − 1 .
5x + 2
When we multiply both sides of this equation by 5x + 2, we see that x must satisfy the equation
5xy + 2y = 3x − 1.
From this equation, we can see that x must satisfy
2y + 1 = x(3 − 5y).
If y = 3/5, this equation has no solution. On the other hand, as long as y ≠ 3/5,
x=
2y + 1
3 − 5y
satisfies this equation. We can conclude that the range of f is y|y ≠ 3/5 .
⎫
⎬
⎭
⎧
⎨
⎩
b. To
find
the
domain
of
f,
we
need
4 − x 2 ≥ 0.
When
we
factor,
we
write
4 − x 2 = (2 − x)(2 + x) ≥ 0. This inequality holds if and only if both terms are positive or both terms
are negative. For both terms to be positive, we need to find x such that
2−x≥0
and
2 + x ≥ 0.
These two inequalities reduce to 2 ≥ x and x ≥ −2. Therefore, the set {x| − 2 ≤ x ≤ 2} must be part
of the domain. For both terms to be negative, we need
2−x≤0
and
2 + x ≥ 0.
These two inequalities also reduce to 2 ≤ x and x ≥ −2. There are no values of x that satisfy both of
these inequalities. Thus, we can conclude the domain of this function is {x| − 2 ≤ x ≤ 2}.
If −2 ≤ x ≤ 2,
⎧
⎨
⎩
1.11
y|0 ≤ y ≤ 2 .
then 0 ≤ 4 − x 2 ≤ 4. Therefore, 0 ≤ 4 − x 2 ≤ 2,
⎫
⎬
⎭
Find the domain and range for the function f (x) = (5x + 2)/(2x − 1).
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and the range of
f is
Chapter 1 | Functions and Graphs
47
The root functions f (x) = x 1/n have defining characteristics depending on whether n is odd or even. For all even integers
n ≥ 2, the domain of f (x) = x 1/n is the interval [0, ∞). For all odd integers n ≥ 1, the domain of f (x) = x 1/n is
the set of all real numbers. Since x 1/n = (−x) 1/n for odd integers n, f (x) = x 1/n is an odd function if n is odd. See the
graphs of root functions for different values of n in Figure 1.21.
n
n
Figure 1.21 (a) If n is even, the domain of f (x) = x is [0, ∞). (b) If n is odd, the domain of f (x) = x is
(−∞, ∞) and the function f (x) = n x is an odd function.
Example 1.17
Finding Domains for Algebraic Functions
For each of the following functions, determine the domain of the function.
3
x2 − 1
a.
f (x) =
b.
f (x) = 2x2+ 5
3x + 4
c.
f (x) = 4 − 3x
d.
f (x) = 2x − 1
3
Solution
a. You cannot divide by zero, so the domain is the set of values x such that x 2 − 1 ≠ 0. Therefore, the
domain is {x|x ≠ ±1}.
b. You need to determine the values of x for which the denominator is zero. Since 3x 2 + 4 ≥ 4 for all real
numbers x, the denominator is never zero. Therefore, the domain is (−∞, ∞).
c. Since the square root of a negative number is not a real number, the domain is the set of values x for
which 4 − 3x ≥ 0. Therefore, the domain is {x|x ≤ 4/3}.
d. The cube root is defined for all real numbers, so the domain is the interval (−∞, ∞).
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Chapter 1 | Functions and Graphs
1.12
Find the domain for each of the following functions: f (x) = (5 − 2x)/(x 2 + 2) and g(x) = 5x − 1.
Transcendental Functions
Thus far, we have discussed algebraic functions. Some functions, however, cannot be described by basic algebraic
operations. These functions are known as transcendental functions because they are said to “transcend,” or go beyond,
algebra. The most common transcendental functions are trigonometric, exponential, and logarithmic functions. A
trigonometric function relates the ratios of two sides of a right triangle. They are sinx, cosx, tanx, cotx, secx, and cscx.
(We discuss trigonometric functions later in the chapter.) An exponential function is a function of the form f (x) = b x,
where the base b > 0, b ≠ 1. A logarithmic function is a function of the form f (x) = log b(x) for some constant
b > 0, b ≠ 1, where log b (x) = y if and only if b y = x. (We also discuss exponential and logarithmic functions later in
the chapter.)
Example 1.18
Classifying Algebraic and Transcendental Functions
Classify each of the following functions, a. through c., as algebraic or transcendental.
a.
3
f (x) = x + 1
4x + 2
b.
f (x) = 2 x
c.
f (x) = sin(2x)
2
Solution
a. Since this function involves basic algebraic operations only, it is an algebraic function.
b. This function cannot be written as a formula that involves only basic algebraic operations, so it is
transcendental. (Note that algebraic functions can only have powers that are rational numbers.)
c. As in part b., this function cannot be written using a formula involving basic algebraic operations only;
therefore, this function is transcendental.
1.13
Is f (x) = x/2 an algebraic or a transcendental function?
Piecewise-Defined Functions
Sometimes a function is defined by different formulas on different parts of its domain. A function with this property is
known as a piecewise-defined function. The absolute value function is an example of a piecewise-defined function because
the formula changes with the sign of x:
⎧−x, x < 0
f (x) = ⎨
.
⎩ x, x ≥ 0
Other piecewise-defined functions may be represented by completely different formulas, depending on the part of the
domain in which a point falls. To graph a piecewise-defined function, we graph each part of the function in its respective
domain, on the same coordinate system. If the formula for a function is different for x < a and x > a, we need to pay
special attention to what happens at x = a when we graph the function. Sometimes the graph needs to include an open or
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Chapter 1 | Functions and Graphs
49
closed circle to indicate the value of the function at x = a. We examine this in the next example.
Example 1.19
Graphing a Piecewise-Defined Function
Sketch a graph of the following piecewise-defined function:
⎧x + 3,
f (x) = ⎨
⎩(x − 2) 2,
x<1
.
x≥1
Solution
Graph the linear function y = x + 3 on the interval (−∞, 1) and graph the quadratic function y = (x − 2) 2
on the interval [1, ∞). Since the value of the function at x = 1 is given by the formula f (x) = (x − 2) 2, we
see that f (1) = 1. To indicate this on the graph, we draw a closed circle at the point (1, 1). The value of the
function is given by f (x) = x + 2 for all x < 1, but not at x = 1. To indicate this on the graph, we draw an
open circle at (1, 4).
Figure 1.22 This piecewise-defined function is linear for
x < 1 and quadratic for x ≥ 1.
1.14
Sketch a graph of the function
⎧2 − x, x ≤ 2
f (x) = ⎨
.
⎩x + 2, x > 2
Example 1.20
Parking Fees Described by a Piecewise-Defined Function
In a big city, drivers are charged variable rates for parking in a parking garage. They are charged $10 for the first
hour or any part of the first hour and an additional $2 for each hour or part thereof up to a maximum of $30 for
50
Chapter 1 | Functions and Graphs
the day. The parking garage is open from 6 a.m. to 12 midnight.
a. Write a piecewise-defined function that describes the cost C to park in the parking garage as a function
of hours parked x.
b. Sketch a graph of this function C(x).
Solution
a. Since the parking garage is open 18 hours each day, the domain for this function is {x|0 < x ≤ 18}. The
cost to park a car at this parking garage can be described piecewise by the function
⎧ 10, 0 < x ≤ 1
⎪ 12, 1 < x ≤ 2
⎪ 14, 2 < x ≤ 3
.
C(x) = ⎨
⎪ 16, 3 < x ≤ 4
⋮
⎪
⎩30, 10 < x ≤ 18
b. The graph of the function consists of several horizontal line segments.
1.15 The cost of mailing a letter is a function of the weight of the letter. Suppose the cost of mailing a letter is
49¢ for the first ounce and 21¢ for each additional ounce. Write a piecewise-defined function describing the
cost C as a function of the weight x for 0 < x ≤ 3, where C is measured in cents and x is measured in
ounces.
Transformations of Functions
We have seen several cases in which we have added, subtracted, or multiplied constants to form variations of simple
functions. In the previous example, for instance, we subtracted 2 from the argument of the function y = x 2 to get the
function f (x) = (x − 2) 2. This subtraction represents a shift of the function y = x 2 two units to the right. A shift,
horizontally or vertically, is a type of transformation of a function. Other transformations include horizontal and vertical
scalings, and reflections about the axes.
A vertical shift of a function occurs if we add or subtract the same constant to each output y. For c > 0, the graph of
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Chapter 1 | Functions and Graphs
51
f (x) + c is a shift of the graph of f (x) up c units, whereas the graph of f (x) − c is a shift of the graph of f (x) down
c units. For example, the graph of the function f (x) = x 3 + 4 is the graph of y = x 3 shifted up 4 units; the graph of the
function f (x) = x 3 − 4 is the graph of y = x 3 shifted down 4 units (Figure 1.23).
Figure 1.23 (a) For c > 0, the graph of y = f (x) + c is a vertical shift up c units of
the graph of y = f (x). (b) For c > 0, the graph of y = f (x) − c is a vertical shift down
c units of the graph of y = f (x).
A horizontal shift of a function occurs if we add or subtract the same constant to each input x. For c > 0, the graph of
f (x + c) is a shift of the graph of f (x) to the left c units; the graph of f (x − c) is a shift of the graph of f (x) to the
right c units. Why does the graph shift left when adding a constant and shift right when subtracting a constant? To answer
this question, let’s look at an example.
Consider the function f (x) = |x + 3| and evaluate this function at x − 3. Since f (x − 3) = |x| and x − 3 < x, the graph
of f (x) = |x + 3| is the graph of y = |x| shifted left 3 units. Similarly, the graph of f (x) = |x − 3| is the graph of y = |x|
shifted right 3 units (Figure 1.24).
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Chapter 1 | Functions and Graphs
Figure 1.24 (a) For c > 0, the graph of y = f (x + c) is a horizontal shift left c units of the graph of y = f (x). (b) For
c > 0, the graph of y = f (x − c) is a horizontal shift right c units of the graph of y = f (x).
A vertical scaling of a graph occurs if we multiply all outputs y of a function by the same positive constant. For c > 0,
the graph of the function c f (x) is the graph of f (x) scaled vertically by a factor of c. If c > 1, the values of the
outputs for the function c f (x) are larger than the values of the outputs for the function f (x); therefore, the graph has been
stretched vertically. If 0 < c < 1, then the outputs of the function c f (x) are smaller, so the graph has been compressed.
For example, the graph of the function f (x) = 3x 2 is the graph of y = x 2 stretched vertically by a factor of 3, whereas the
graph of f (x) = x 2 /3 is the graph of y = x 2 compressed vertically by a factor of 3 (Figure 1.25).
Figure 1.25 (a) If c > 1, the graph of y = c f (x) is a vertical stretch of the graph
of y = f (x). (b) If 0 < c < 1, the graph of y = c f (x) is a vertical compression of
the graph of y = f (x).
The horizontal scaling of a function occurs if we multiply the inputs x by the same positive constant. For c > 0, the
graph of the function f (cx) is the graph of f (x) scaled horizontally by a factor of c. If c > 1, the graph of f (cx) is the
graph of f (x) compressed horizontally. If 0 < c < 1, the graph of f (cx) is the graph of f (x) stretched horizontally. For
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Chapter 1 | Functions and Graphs
53
example, consider the function f (x) = 2x and evaluate f at x/2. Since f (x/2) = x, the graph of f (x) = 2x is the
graph of y = x compressed horizontally. The graph of y = x/2 is a horizontal stretch of the graph of y = x (Figure
1.26).
Figure 1.26 (a) If c > 1, the graph of y = f (cx) is a horizontal compression of the graph
of y = f (x). (b) If 0 < c < 1, the graph of y = f (cx) is a horizontal stretch of the graph of
y = f (x).
We have explored what happens to the graph of a function f when we multiply f by a constant c > 0 to get a new
function c f (x). We have also discussed what happens to the graph of a function f when we multiply the independent
variable x by c > 0 to get a new function f (cx). However, we have not addressed what happens to the graph of the
function if the constant c is negative. If we have a constant c < 0, we can write c as a positive number multiplied by
−1; but, what kind of transformation do we get when we multiply the function or its argument by −1 ? When we multiply
all the outputs by −1, we get a reflection about the x -axis. When we multiply all inputs by −1, we get a reflection
about the y -axis. For example, the graph of f (x) = −(x 3 + 1) is the graph of y = (x 3 + 1) reflected about the x -axis.
The graph of f (x) = (−x) 3 + 1 is the graph of y = x 3 + 1 reflected about the y -axis (Figure 1.27).
54
Chapter 1 | Functions and Graphs
Figure 1.27 (a) The graph of y = − f (x) is the graph of
y = f (x) reflected about the x -axis. (b) The graph of
y = f (−x) is the graph of y = f (x) reflected about the
y -axis.
If the graph of a function consists of more than one transformation of another graph, it is important to transform the graph
in the correct order. Given a function f (x), the graph of the related function y = c f ⎛⎝a(x + b)⎞⎠ + d can be obtained from
the graph of y = f (x) by performing the transformations in the following order.
1. Horizontal shift of the graph of y = f (x). If b > 0, shift left. If b < 0, shift right.
2. Horizontal scaling of the graph of y = f (x + b) by a factor of |a|. If a < 0, reflect the graph about the y -axis.
3. Vertical scaling of the graph of y = f (a(x + b)) by a factor of |c|. If c < 0, reflect the graph about the x -axis.
4. Vertical shift of the graph of y = c f (a(x + b)). If d > 0, shift up. If d < 0, shift down.
We can summarize the different transformations and their related effects on the graph of a function in the following table.
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Chapter 1 | Functions and Graphs
Transformation of f(c > 0)
55
Effect on the graph of f
f (x) + c
Vertical shift up c units
f (x) − c
Vertical shift down c units
f (x + c)
Shift left by c units
f (x − c)
Shift right by c units
Vertical stretch if c > 1;
c f (x)
vertical compression if 0 < c < 1
f (cx)
Horizontal stretch if 0 < c < 1; horizontal compression if c > 1
− f (x)
Reflection about the x -axis
f (−x)
Reflection about the y -axis
Table 1.7 Transformations of Functions
Example 1.21
Transforming a Function
For each of the following functions, a. and b., sketch a graph by using a sequence of transformations of a wellknown function.
a.
f (x) = −|x + 2| − 3
b.
f (x) = 3 −x + 1
Solution
a. Starting with the graph of y = |x|, shift 2 units to the left, reflect about the x -axis, and then shift down
3 units.
56
Chapter 1 | Functions and Graphs
Figure 1.28 The function f (x) = −|x + 2| − 3 can be
viewed as a sequence of three transformations of the function
y = |x|.
b. Starting with the graph of y = x, reflect about the y -axis, stretch the graph vertically by a factor of 3,
and move up 1 unit.
Figure 1.29 The function f (x) = 3 −x + 1 can be viewed
as a sequence of three transformations of the function y = x.
1.16
Describe how the function f (x) = −(x + 1) 2 − 4 can be graphed using the graph of y = x 2 and a
sequence of transformations.
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Chapter 1 | Functions and Graphs
57
1.2 EXERCISES
For the following exercises, for each pair of points, a.
find the slope of the line passing through the points and
b. indicate whether the line is increasing, decreasing,
horizontal, or vertical.
59. (−2, 4) and (1, 1)
60. (−1, 4) and (3, −1)
79. 4y + 24 = 0
80. 8x − 4 = 0
81. 2x + 3y = 6
82. 6x − 5y + 15 = 0
61. (3, 5) and (−1, 2)
For the following exercises, for each polynomial, a. find the
degree; b. find the zeros, if any; c. find the y -intercept(s),
62. (6, 4) and (4, −3)
if any; d. use the leading coefficient to determine the
graph’s end behavior; and e. determine algebraically
whether the polynomial is even, odd, or neither.
63. (2, 3) and (5, 7)
83.
f (x) = 2x 2 − 3x − 5
65. (2, 4) and (1, 4)
84.
f (x) = −3x 2 + 6x
66. (1, 4) and (1, 0)
85.
f (x) = 1 x 2 − 1
2
For the following exercises, write the equation of the line
satisfying the given conditions in slope-intercept form.
86.
f (x) = x 3 + 3x 2 − x − 3
67. Slope = −6, passes through (1, 3)
87.
f (x) = 3x − x 3
68. Slope = 3, passes through (−3, 2)
For the following exercises, use the graph of f (x) = x 2 to
69. Slope = 1 , passes through (0, 4)
graph each transformed function g.
64. (1, 9) and (−8, 5)
3
70. Slope = 2 , x -intercept = 8
5
71. Passing through (2, 1) and (−2, −1)
72. Passing through (−3, 7) and (1, 2)
88. g(x) = x 2 − 1
89. g(x) = (x + 3) 2 + 1
For the following exercises, use the graph of f (x) = x to
graph each transformed function g.
73. x -intercept = 5 and y -intercept = −3
90. g(x) = x + 2
74. x -Intercept = −6 and y -intercept = 9
91. g(x) = − x − 1
For the following exercises, for each linear equation, a. give
the slope m and y -intercept b, if any, and b. graph the line.
For the following exercises, use the graph of y = f (x) to
75. y = 2x − 3
76. y = − 1 x + 1
7
77.
f (x) = −6x
78.
f (x) = −5x + 4
graph each transformed function g.
58
Chapter 1 | Functions and Graphs
103. [T] A company purchases some computer equipment
for $20,500. At the end of a 3-year period, the value of the
equipment has decreased linearly to $12,300.
a. Find a function y = V(t) that determines the value
V of the equipment at the end of t years.
b. Find and interpret the meaning of the x - and y
-intercepts for this situation.
c. What is the value of the equipment at the end of 5
years?
d. When will the value of the equipment be $3000?
104. [T] Total online shopping during the Christmas
holidays has increased dramatically during the past 5 years.
In 2012 (t = 0), total online holiday sales were $42.3
billion, whereas in 2013 they were $48.1 billion.
a. Find a linear function S that estimates the total
online holiday sales in the year t.
b. Interpret the slope of the graph of S.
c. Use part a. to predict the year when online shopping
during Christmas will reach $60 billion.
92. g(x) = f (x) + 1
93. g(x) = f (x − 1) + 2
For the following exercises, for each of the piecewisedefined functions, a. evaluate at the given values of the
independent variable and b. sketch the graph.
94.
⎧ 4x + 3, x ≤ 0
f (x) = ⎨
; f (−3); f (0); f (2)
⎩−x + 1, x > 0
95.
⎧x 2 − 3, x < 0
f (x) = ⎨
; f (−4); f (0); f (2)
⎩4x − 3, x ≥ 0
⎧x
96. h(x) = ⎨
⎩
+ 1, x ≤ 5
; h(0); h(π); h(5)
4, x > 5
⎧ 3 ,x≠2
; g(0); g(−4); g(2)
97. g(x) = ⎨ x − 2
⎩4, x = 2
For the following exercises, determine whether the
statement is true or false. Explain why.
98.
f (x) = (4x + 1)/(7x − 2)
is
a
transcendental
function.
99. g(x) = 3 x is an odd root function
100. A logarithmic function is an algebraic function.
101. A function of the form f (x) = x b, where b is a
real valued constant, is an exponential function.
102. The domain of an even root function is all real
numbers.
105. [T] A family bakery makes cupcakes and sells them
at local outdoor festivals. For a music festival, there is a
fixed cost of $125 to set up a cupcake stand. The owner
estimates that it costs $0.75 to make each cupcake. The
owner is interested in determining the total cost C as a
function of number of cupcakes made.
a. Find a linear function that relates cost C to x, the
number of cupcakes made.
b. Find the cost to bake 160 cupcakes.
c. If the owner sells the cupcakes for $1.50 apiece,
how many cupcakes does she need to sell to start
making profit? (Hint: Use the INTERSECTION
function on a calculator to find this number.)
106. [T] A house purchased for $250,000 is expected to
be worth twice its purchase price in 18 years.
a. Find a linear function that models the price P of
the house versus the number of years t since the
original purchase.
b. Interpret the slope of the graph of P.
c. Find the price of the house 15 years from when it
was originally purchased.
107. [T] A car was purchased for $26,000. The value of
the car depreciates by $1500 per year.
a. Find a linear function that models the value V of the
car after t years.
b. Find and interpret V(4).
108. [T] A condominium in an upscale part of the city was
purchased for $432,000. In 35 years it is worth $60,500.
Find the rate of depreciation.
109. [T] The total cost C (in thousands of dollars) to
produce a certain item is modeled by the function
C(x) = 10.50x + 28,500, where x is the number of items
produced. Determine the cost to produce 175 items.
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Chapter 1 | Functions and Graphs
110. [T] A professor asks her class to report the amount
of time t they spent writing two assignments. Most students
report that it takes them about 45 minutes to type a fourpage assignment and about 1.5 hours to type a nine-page
assignment.
a. Find the linear function y = N(t) that models this
situation, where N is the number of pages typed
and t is the time in minutes.
b. Use part a. to determine how many pages can be
typed in 2 hours.
c. Use part a. to determine how long it takes to type a
20-page assignment.
111. [T] The output (as a percent of total capacity) of
nuclear power plants in the United States can be modeled
by the function P(t) = 1.8576t + 68.052, where t is time
in years and t = 0 corresponds to the beginning of 2000.
Use the model to predict the percentage output in 2015.
112. [T] The admissions office at a public university
estimates that 65% of the students offered admission to the
class of 2019 will actually enroll.
a. Find the linear function y = N(x), where N is
the number of students that actually enroll and x is
the number of all students offered admission to the
class of 2019.
b. If the university wants the 2019 freshman class size
to be 1350, determine how many students should be
admitted.
59
60
Chapter 1 | Functions and Graphs
1.3 | Trigonometric Functions
Learning Objectives
1.3.1
1.3.2
1.3.3
1.3.4
1.3.5
Convert angle measures between degrees and radians.
Recognize the triangular and circular definitions of the basic trigonometric functions.
Write the basic trigonometric identities.
Identify the graphs and periods of the trigonometric functions.
Describe the shift of a sine or cosine graph from the equation of the function.
Trigonometric functions are used to model many phenomena, including sound waves, vibrations of strings, alternating
electrical current, and the motion of pendulums. In fact, almost any repetitive, or cyclical, motion can be modeled by some
combination of trigonometric functions. In this section, we define the six basic trigonometric functions and look at some of
the main identities involving these functions.
Radian Measure
To use trigonometric functions, we first must understand how to measure the angles. Although we can use both radians and
degrees, radians are a more natural measurement because they are related directly to the unit circle, a circle with radius 1.
The radian measure of an angle is defined as follows. Given an angle θ, let s be the length of the corresponding arc on
the unit circle (Figure 1.30). We say the angle corresponding to the arc of length 1 has radian measure 1.
Figure 1.30 The radian measure of an angle θ is the arc
length s of the associated arc on the unit circle.
Since an angle of 360° corresponds to the circumference of a circle, or an arc of length 2π, we conclude that an angle
with a degree measure of 360° has a radian measure of 2π. Similarly, we see that 180° is equivalent to π radians. Table
1.8 shows the relationship between common degree and radian values.
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Chapter 1 | Functions and Graphs
61
Degrees
Radians
Degrees
Radians
0
0
120
2π/3
30
π/6
135
3π/4
45
π/4
150
5π/6
60
π/3
180
π
90
π/2
Table 1.8 Common Angles Expressed in Degrees and
Radians
Example 1.22
Converting between Radians and Degrees
a. Express 225° using radians.
b. Express 5π/3 rad using degrees.
Solution
Use the fact that 180° is equivalent to π radians as a conversion factor: 1 = π rad = 180° .
180°
a.
225° = 225° · π = 5π rad
180°
4
b.
5π rad = 5π · 180° = 300°
π
3
3
1.17
π rad
Express 210° using radians. Express 11π/6 rad using degrees.
The Six Basic Trigonometric Functions
Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any
circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the
sides and angles of a triangle.
To define the trigonometric functions, first consider the unit circle centered at the origin and a point P = (x, y) on the unit
circle. Let θ be an angle with an initial side that lies along the positive x -axis and with a terminal side that is the line
segment OP. An angle in this position is said to be in standard position (Figure 1.31). We can then define the values of
the six trigonometric functions for θ in terms of the coordinates x and y.
62
Chapter 1 | Functions and Graphs
Figure 1.31 The angle θ is in standard position. The values
of the trigonometric functions for θ are defined in terms of the
coordinates x and y.
Definition
Let P = (x, y) be a point on the unit circle centered at the origin O. Let θ be an angle with an initial side along the
positive x -axis and a terminal side given by the line segment OP. The trigonometric functions are then defined as
sinθ = y
cosθ = x
y
tanθ = x
cscθ = 1y
secθ = 1x
cot θ = xy
(1.9)
If x = 0, secθ and tanθ are undefined. If y = 0, then cot θ and cscθ are undefined.
We can see that for a point P = (x, y) on a circle of radius r with a corresponding angle θ, the coordinates x and y
satisfy
cosθ = xr
x = r cosθ
y
sinθ = r
y = r sinθ.
The values of the other trigonometric functions can be expressed in terms of x, y, and r (Figure 1.32).
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Chapter 1 | Functions and Graphs
63
Figure 1.32 For a point P = (x, y) on a circle of radius r,
the coordinates x and y satisfy x = r cosθ and y = r sinθ.
Table 1.9 shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine
the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric
functions are calculated easily from the values of sinθ and cosθ.
θ
sinθ
cosθ
0
0
1
π
6
1
2
3
2
π
4
2
2
2
2
π
3
3
2
1
2
π
2
1
0
Table 1.9 Values of sinθ
and cosθ at Major Angles
θ in the First Quadrant
Example 1.23
Evaluating Trigonometric Functions
Evaluate each of the following expressions.
⎛
⎞
a.
sin⎝2π ⎠
3
b.
cos⎝− 5π ⎠
6
⎛
⎞
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Chapter 1 | Functions and Graphs
c.
⎛
⎞
tan⎝15π ⎠
4
Solution
⎛
⎞
⎛ ⎞
a. On the unit circle, the angle θ = 2π corresponds to the point ⎝− 1 , 3 ⎠. Therefore, sin⎝2π ⎠ = y = 3 .
3
3
2
2 2
b. An angle θ = − 5π corresponds to a revolution in the negative direction, as shown. Therefore,
6
⎞
⎛
cos⎝− 5π ⎠ = x = − 3 .
2
6
c. An angle θ = 15π = 2π + 7π . Therefore, this angle corresponds to more than one revolution, as shown.
4
4
⎛
⎞
Knowing the fact that an angle of 7π corresponds to the point ⎝ 2 , − 2 ⎠, we can conclude that
4
2
2
y
⎛
⎞
tan⎝15π ⎠ = x = −1.
4
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Chapter 1 | Functions and Graphs
1.18
65
Evaluate cos(3π/4) and sin(−π/6).
As mentioned earlier, the ratios of the side lengths of a right triangle can be expressed in terms of the trigonometric functions
evaluated at either of the acute angles of the triangle. Let θ be one of the acute angles. Let A be the length of the adjacent
leg, O be the length of the opposite leg, and H be the length of the hypotenuse. By inscribing the triangle into a circle of
radius H, as shown in Figure 1.33, we see that A, H, and O satisfy the following relationships with θ:
sinθ = O
H
cosθ = A
H
tanθ = O
A
cscθ = H
O
H
secθ =
A
cot θ = A
O
Figure 1.33 By inscribing a right triangle in a circle, we can
express the ratios of the side lengths in terms of the
trigonometric functions evaluated at θ.
Example 1.24
Constructing a Wooden Ramp
A wooden ramp is to be built with one end on the ground and the other end at the top of a short staircase. If the
top of the staircase is 4 ft from the ground and the angle between the ground and the ramp is to be 10°, how
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Chapter 1 | Functions and Graphs
long does the ramp need to be?
Solution
Let x denote the length of the ramp. In the following image, we see that x needs to satisfy the equation
sin(10°) = 4/x. Solving this equation for x, we see that x = 4/sin(10°) ≈ 23.035 ft.
1.19 A house painter wants to lean a 20 -ft ladder against a house. If the angle between the base of the ladder
and the ground is to be 60°, how far from the house should she place the base of the ladder?
Trigonometric Identities
A trigonometric identity is an equation involving trigonometric functions that is true for all angles θ for which the
functions are defined. We can use the identities to help us solve or simplify equations. The main trigonometric identities are
listed next.
Rule: Trigonometric Identities
Reciprocal identities
tanθ = sinθ
cosθ
cscθ = 1
sinθ
cot θ = cosθ
sinθ
secθ = 1
cosθ
Pythagorean identities
sin 2 θ + cos 2 θ = 1
1 + tan 2 θ = sec 2 θ
1 + cot 2 θ = csc 2 θ
Addition and subtraction formulas
sin⎛⎝α ± β⎞⎠ = sinαcos β ± cosαsin β
cos(α ± β) = cosαcos β ∓ sinαsin β
Double-angle formulas
sin(2θ) = 2sinθ cosθ
cos(2θ) = 2cos 2 θ − 1 = 1 − 2sin 2 θ = cos 2 θ − sin 2 θ
Example 1.25
Solving Trigonometric Equations
For each of the following equations, use a trigonometric identity to find all solutions.
a.
1 + cos(2θ) = cosθ
b.
sin(2θ) = tanθ
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Chapter 1 | Functions and Graphs
67
Solution
a. Using the double-angle formula for cos(2θ), we see that θ is a solution of
1 + cos(2θ) = cosθ
if and only if
1 + 2cos 2 θ − 1 = cosθ,
which is true if and only if
2cos 2 θ − cosθ = 0.
To solve this equation, it is important to note that we need to factor the left-hand side and not divide both
sides of the equation by cosθ. The problem with dividing by cosθ is that it is possible that cosθ is
zero. In fact, if we did divide both sides of the equation by cosθ, we would miss some of the solutions
of the original equation. Factoring the left-hand side of the equation, we see that θ is a solution of this
equation if and only if
cosθ(2cosθ − 1) = 0.
Since cosθ = 0 when
θ = π , π ± π, π ± 2π,…,
2 2
2
and cosθ = 1/2 when
θ = π , π ± 2π,… or θ = − π , − π ± 2π,…,
3 3
3
3
we conclude that the set of solutions to this equation is
θ = π + nπ, θ = π + 2nπ, and θ = − π + 2nπ, n = 0, ± 1, ± 2,….
3
2
3
b. Using the double-angle formula for sin(2θ) and the reciprocal identity for tan(θ), the equation can be
written as
2sinθ cosθ = sinθ .
cosθ
To solve this equation, we multiply both sides by cosθ to eliminate the denominator, and say that if θ
satisfies this equation, then θ satisfies the equation
2sinθcos 2 θ − sinθ = 0.
However, we need to be a little careful here. Even if θ satisfies this new equation, it may not satisfy the
original equation because, to satisfy the original equation, we would need to be able to divide both sides
of the equation by cosθ. However, if cosθ = 0, we cannot divide both sides of the equation by cosθ.
Therefore, it is possible that we may arrive at extraneous solutions. So, at the end, it is important to check
for extraneous solutions. Returning to the equation, it is important that we factor sinθ out of both terms
on the left-hand side instead of dividing both sides of the equation by sinθ. Factoring the left-hand side
of the equation, we can rewrite this equation as
sinθ(2cos 2 θ − 1) = 0.
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Chapter 1 | Functions and Graphs
Therefore, the solutions are given by the angles θ such that sinθ = 0 or cos 2 θ = 1/2. The solutions
of the first equation are θ = 0, ± π, ± 2π,…. The solutions of the second equation are
θ = π/4, (π/4) ± (π/2), (π/4) ± π,…. After checking for extraneous solutions, the set of solutions to the
equation is
θ = nπ
1.20
and θ = π + nπ , n = 0, ± 1, ± 2,….
4
2
Find all solutions to the equation cos(2θ) = sinθ.
Example 1.26
Proving a Trigonometric Identity
Prove the trigonometric identity 1 + tan 2 θ = sec 2 θ.
Solution
We start with the identity
sin 2 θ + cos 2 θ = 1.
Dividing both sides of this equation by cos 2 θ, we obtain
sin 2 θ + 1 = 1 .
cos 2 θ
cos 2 θ
Since sinθ/cosθ = tanθ and 1/cosθ = secθ, we conclude that
tan 2 θ + 1 = sec 2 θ.
1.21
Prove the trigonometric identity 1 + cot 2 θ = csc 2 θ.
Graphs and Periods of the Trigonometric Functions
We have seen that as we travel around the unit circle, the values of the trigonometric functions repeat. We can see this
pattern in the graphs of the functions. Let P = (x, y) be a point on the unit circle and let θ be the corresponding angle
. Since the angle θ and θ + 2π correspond to the same point P, the values of the trigonometric functions at θ and
at θ + 2π are the same. Consequently, the trigonometric functions are periodic functions. The period of a function f is
defined to be the smallest positive value p such that f (x + p) = f (x) for all values x in the domain of f . The sine,
cosine, secant, and cosecant functions have a period of 2π. Since the tangent and cotangent functions repeat on an interval
of length π, their period is π (Figure 1.34).
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69
Figure 1.34 The six trigonometric functions are periodic.
Just as with algebraic functions, we can apply transformations to trigonometric functions. In particular, consider the
following function:
f (x) = Asin⎛⎝B(x − α)⎞⎠ + C.
(1.10)
In Figure 1.35, the constant α causes a horizontal or phase shift. The factor B changes the period. This transformed
sine function will have a period 2π/|B|. The factor A results in a vertical stretch by a factor of | A|. We say | A| is the
“amplitude of f . ” The constant C causes a vertical shift.
Figure 1.35 A graph of a general sine function.
Notice in Figure 1.34 that the graph of y = cos x is the graph of y = sin x shifted to the left π/2 units. Therefore, we
70
Chapter 1 | Functions and Graphs
can write cos x = sin(x + π/2). Similarly, we can view the graph of y = sin x as the graph of y = cos x shifted right π/2
units, and state that sin x = cos(x − π/2).
A shifted sine curve arises naturally when graphing the number of hours of daylight in a given location as a function of
the day of the year. For example, suppose a city reports that June 21 is the longest day of the year with 15.7 hours and
December 21 is the shortest day of the year with 8.3 hours. It can be shown that the function
⎛
⎞
h(t) = 3.7sin⎝ 2π (x − 80.5)⎠ + 12
365
is a model for the number of hours of daylight h as a function of day of the year t (Figure 1.36).
Figure 1.36 The hours of daylight as a function of day of the year can be modeled
by a shifted sine curve.
Example 1.27
Sketching the Graph of a Transformed Sine Curve
⎛ ⎛
⎞⎞
Sketch a graph of f (x) = 3sin⎝2⎝x − π ⎠⎠ + 1.
4
Solution
This graph is a phase shift of y = sin(x) to the right by π/4 units, followed by a horizontal compression by a
factor of 2, a vertical stretch by a factor of 3, and then a vertical shift by 1 unit. The period of f is π.
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Chapter 1 | Functions and Graphs
1.22
Describe the relationship between the graph of f (x) = 3sin(4x) − 5 and the graph of y = sin(x).
71
72
Chapter 1 | Functions and Graphs
1.3 EXERCISES
For the following exercises, convert each angle in degrees
to radians. Write the answer as a multiple of π.
113. 240°
114. 15°
115. −60°
116. −225°
129. a = 4, c = 7
117. 330°
130. a = 21, c = 29
For the following exercises, convert each angle in radians
to degrees.
131. a = 85.3, b = 125.5
118.
π rad
2
133. a = 84, b = 13
119.
7π rad
6
120.
11π rad
2
134. b = 28, c = 35
For the following exercises, P is a point on the unit circle.
a. Find the (exact) missing coordinate value of each point
and b. find the values of the six trigonometric functions for
the angle θ with a terminal side that passes through point
P. Rationalize denominators.
121. −3π rad
122.
5π rad
12
⎛
Evaluate the following functional values.
⎛
3
⎛
17
⎞
⎛
⎞
137. P⎝x, 7 ⎠, x < 0
3
⎞
⎞
⎛
138. P⎝x, − 15 ⎠, x > 0
4
125. sin⎝− 3π ⎠
4
⎞
136. P⎝−15 , y⎠, y < 0
124. tan⎝19π ⎠
4
⎞
135. P⎝ 7 , y⎠, y > 0
25
⎛
⎞
123. cos⎝4π ⎠
⎛
132. b = 40, c = 41
⎛ ⎞
126. sec⎝π ⎠
⎛
⎞
For the following exercises, simplify each expression by
writing it in terms of sines and cosines, then simplify. The
final answer does not have to be in terms of sine and cosine
only.
⎛
⎞
139. tan 2 x + sin x csc x
6
127. sin⎝ π ⎠
12
128. cos⎝5π ⎠
12
For the following exercises, consider triangle ABC, a right
triangle with a right angle at C. a. Find the missing side of
the triangle. b. Find the six trigonometric function values
for the angle at A. Where necessary, round to one decimal
place.
140. sec x sin x cot x
141.
tan 2 x
sec 2 x
142. sec x − cos x
143. (1 + tanθ) 2 − 2tanθ
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Chapter 1 | Functions and Graphs
144. sin x(csc x − sin x)
145.
cost + sint
sint 1 + cost
146.
1 + tan 2 α
1 + cot 2 α
73
163.
For the following exercises, verify that each equation is an
identity.
147.
tanθ cot θ = sinθ
cscθ
148.
sec 2 θ = secθ cscθ
tanθ
149.
sint cost
csct + sect = 1
150.
sin x + cos x − 1 = 0
cos x + 1
sin x
164.
151. cot γ + tanγ = secγ cscγ
152. sin 2 β + tan 2 β + cos 2 β = sec 2 β
153.
1
1
+
= 2sec 2 α
1 − sinα 1 + sinα
154.
tanθ − cot θ = sec 2 θ − csc 2 θ
sinθ cosθ
For the following exercises, solve the trigonometric
equations on the interval 0 ≤ θ < 2π.
155. 2sinθ − 1 = 0
156. 1 + cosθ = 1
2
157. 2tan 2 θ = 2
158. 4sin 2 θ − 2 = 0
159.
3cot θ + 1 = 0
160. 3secθ − 2 3 = 0
161. 2cosθ sinθ = sinθ
162. csc 2 θ + 2cscθ + 1 = 0
For the following exercises, each graph is of the form
y = Asin Bx or y = Acos Bx, where B > 0. Write the
equation of the graph.
165.
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Chapter 1 | Functions and Graphs
166.
175. [T] As a point P moves around a circle, the measure
of the angle changes. The measure of how fast the angle
is changing is called angular speed, ω, and is given by
ω = θ/t, where θ is in radians and t is time. Find the
angular speed for the given data. Round to the nearest
θ = 7π rad, t = 10
sec
b.
4
θ = 3π rad, t = 8 sec c. θ = 2π rad, t = 1 min d.
9
5
θ = 23.76rad,t = 14 min
thousandth.
For the following exercises, find a. the amplitude, b. the
period, and c. the phase shift with direction for each
function.
⎛
⎞
167. y = sin⎝x − π ⎠
4
168. y = 3cos(2x + 3)
⎛ ⎞
169. y = −1 sin⎝1 x⎠
2
4
⎛
⎞
170. y = 2cos⎝x − π ⎠
3
171. y = −3sin(πx + 2)
⎛
⎞
172. y = 4cos⎝2x − π ⎠
2
173. [T] The diameter of a wheel rolling on the ground
is 40 in. If the wheel rotates through an angle of 120°,
how many inches does it move? Approximate to the nearest
whole inch.
174. [T] Find the length of the arc intercepted by central
angle θ in a circle of radius r. Round to the nearest
hundredth. a. r = 12.8 cm, θ = 5π rad b. r = 4.378 cm,
6
θ = 7π rad c. r = 0.964 cm, θ = 50° d. r = 8.55 cm,
6
θ = 325°
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a.
176. [T] A total of 250,000 m2 of land is needed to build a
nuclear power plant. Suppose it is decided that the area on
which the power plant is to be built should be circular.
a. Find the radius of the circular land area.
b. If the land area is to form a 45° sector of a circle
instead of a whole circle, find the length of the
curved side.
177. [T] The area of an isosceles triangle with equal sides
of length x is 1 x 2 sinθ, where θ is the angle formed by
2
the two sides. Find the area of an isosceles triangle with
equal sides of length 8 in. and angle θ = 5π/12 rad.
178. [T] A particle travels in a circular path at a constant
angular speed ω. The angular speed is modeled by the
function ω = 9|cos(πt − π/12)|. Determine the angular
speed at t = 9 sec.
179. [T] An alternating current for outlets in a home has
voltage given by the function V(t) = 150cos368t, where
V is the voltage in volts at time t in seconds.
a. Find the period of the function and interpret its
meaning.
b. Determine the number of periods that occur when 1
sec has passed.
180. [T] The number of hours of daylight in a northeast
city is modeled by the function
⎡
⎤
N(t) = 12 + 3sin⎣ 2π (t − 79)⎦,
365
where t is the number of days after January 1.
a. Find the amplitude and period.
b. Determine the number of hours of daylight on the
longest day of the year.
c. Determine the number of hours of daylight on the
shortest day of the year.
d. Determine the number of hours of daylight 90 days
after January 1.
e. Sketch the graph of the function for one period
starting on January 1.
Chapter 1 | Functions and Graphs
75
⎡
⎤
181. [T] Suppose that T = 50 + 10sin⎣ π (t − 8)⎦ is a
12
mathematical model of the temperature (in degrees
Fahrenheit) at t hours after midnight on a certain day of the
week.
a. Determine the amplitude and period.
b. Find the temperature 7 hours after midnight.
c. At what time does T = 60° ?
d. Sketch the graph of T over 0 ≤ t ≤ 24.
⎛
⎞
182. [T] The function H(t) = 8sin⎝π t⎠ models the height
6
H (in feet) of the tide t hours after midnight. Assume that
t = 0 is midnight.
a. Find the amplitude and period.
b. Graph the function over one period.
c. What is the height of the tide at 4:30 a.m.?
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Chapter 1 | Functions and Graphs
1.4 | Inverse Functions
Learning Objectives
1.4.1
1.4.2
1.4.3
1.4.4
1.4.5
Determine the conditions for when a function has an inverse.
Use the horizontal line test to recognize when a function is one-to-one.
Find the inverse of a given function.
Draw the graph of an inverse function.
Evaluate inverse trigonometric functions.
An inverse function reverses the operation done by a particular function. In other words, whatever a function does, the
inverse function undoes it. In this section, we define an inverse function formally and state the necessary conditions for an
inverse function to exist. We examine how to find an inverse function and study the relationship between the graph of a
function and the graph of its inverse. Then we apply these ideas to define and discuss properties of the inverse trigonometric
functions.
Existence of an Inverse Function
We begin with an example. Given a function f and an output y = f (x),
we are often interested in finding what
value or values x were mapped to y by f . For example, consider the function f (x) = x 3 + 4. Since any output
3
y = x 3 + 4, we can solve this equation for x to find that the input is x = y − 4. This equation defines x as a function
3
of y. Denoting this function as f −1, and writing x = f −1 (y) = y − 4, we see that for any x in the domain of
f , f −1 ⎛⎝ f (x)⎞⎠ = f −1 ⎛⎝x 3 + 4⎞⎠ = x. Thus, this new function, f −1, “undid” what the original function f did. A function
with this property is called the inverse function of the original function.
Definition
Given a function f with domain D and range R, its inverse function (if it exists) is the function f −1 with domain
R and range D such that f −1 (y) = x if f (x) = y. In other words, for a function f and its inverse f −1,
f −1 ⎛⎝ f (x)⎞⎠ = x for all x in D, and f ⎛⎝ f −1 (y)⎞⎠ = y for all y in R.
(1.11)
Note that f −1 is read as “f inverse.” Here, the −1 is not used as an exponent and f −1 (x) ≠ 1/ f (x). Figure 1.37 shows
the relationship between the domain and range of f and the domain and range of f −1.
Figure 1.37 Given a function f and its inverse
f −1, f −1 (y) = x if and only if f (x) = y. The range of f
becomes the domain of f −1 and the domain of f becomes the
range of f −1.
Recall that a function has exactly one output for each input. Therefore, to define an inverse function, we need to map each
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Chapter 1 | Functions and Graphs
77
input to exactly one output. For example, let’s try to find the inverse function for f (x) = x 2. Solving the equation y = x 2
for x, we arrive at the equation x = ± y. This equation does not describe x as a function of y because there are two
solutions to this equation for every y > 0. The problem with trying to find an inverse function for f (x) = x 2 is that two
inputs are sent to the same output for each output y > 0. The function f (x) = x 3 + 4 discussed earlier did not have this
problem. For that function, each input was sent to a different output. A function that sends each input to a different output
is called a one-to-one function.
Definition
We say a f is a one-to-one function if f (x 1) ≠ f (x 2) when x 1 ≠ x 2.
One way to determine whether a function is one-to-one is by looking at its graph. If a function is one-to-one, then no two
inputs can be sent to the same output. Therefore, if we draw a horizontal line anywhere in the xy -plane, according to the
horizontal line test, it cannot intersect the graph more than once. We note that the horizontal line test is different from
the vertical line test. The vertical line test determines whether a graph is the graph of a function. The horizontal line test
determines whether a function is one-to-one (Figure 1.38).
Rule: Horizontal Line Test
A function f is one-to-one if and only if every horizontal line intersects the graph of f no more than once.
Figure 1.38 (a) The function f (x) = x 2 is not one-to-one
because it fails the horizontal line test. (b) The function
f (x) = x 3 is one-to-one because it passes the horizontal line
test.
Example 1.28
Determining Whether a Function Is One-to-One
For each of the following functions, use the horizontal line test to determine whether it is one-to-one.
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Chapter 1 | Functions and Graphs
a.
b.
Solution
a. Since the horizontal line y = n for any integer n ≥ 0 intersects the graph more than once, this function
is not one-to-one.
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Chapter 1 | Functions and Graphs
b. Since every horizontal line intersects the graph once (at most), this function is one-to-one.
1.23
Is the function f graphed in the following image one-to-one?
79
80
Chapter 1 | Functions and Graphs
Finding a Function’s Inverse
We can now consider one-to-one functions and show how to find their inverses. Recall that a function maps elements in
the domain of f to elements in the range of f . The inverse function maps each element from the range of f back to its
corresponding element from the domain of f . Therefore, to find the inverse function of a one-to-one function f , given
any y in the range of f , we need to determine which x in the domain of f satisfies f (x) = y. Since f is one-to-one,
there is exactly one such value x. We can find that value x by solving the equation f (x) = y for x. Doing so, we are
able to write x as a function of y where the domain of this function is the range of f and the range of this new function
is the domain of f . Consequently, this function is the inverse of f , and we write x = f −1(y). Since we typically use the
variable x to denote the independent variable and y to denote the dependent variable, we often interchange the roles of x
and y, and write y = f −1(x). Representing the inverse function in this way is also helpful later when we graph a function
f and its inverse f −1 on the same axes.
Problem-Solving Strategy: Finding an Inverse Function
1. Solve the equation y = f (x) for x.
2. Interchange the variables x and y and write y = f −1(x).
Example 1.29
Finding an Inverse Function
Find the inverse for the function f (x) = 3x − 4. State the domain and range of the inverse function. Verify that
f −1( f (x)) = x.
Solution
Follow the steps outlined in the strategy.
Step 1. If y = 3x − 4, then 3x = y + 4 and x = 1 y + 4 .
3
3
Step 2. Rewrite as y = 1 x + 4 and let y = f −1 (x).
3
3
Therefore, f −1 (x) = 1 x + 4 .
3
3
Since the domain of f is (−∞, ∞), the range of f −1 is (−∞, ∞). Since the range of f is (−∞, ∞), the
domain of f −1 is (−∞, ∞).
You can verify that f −1( f (x)) = x by writing
f −1( f (x)) = f −1(3x − 4) = 1 (3x − 4) + 4 = x − 4 + 4 = x.
3
3
3 3
Note that for f −1(x) to be the inverse of f (x), both f −1( f (x)) = x and f ( f −1(x)) = x for all x in the domain
of the inside function.
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Chapter 1 | Functions and Graphs
1.24
81
Find the inverse of the function f (x) = 3x/(x − 2). State the domain and range of the inverse function.
Graphing Inverse Functions
Let’s consider the relationship between the graph of a function f and the graph of its inverse. Consider the graph of f
shown in Figure 1.39 and a point (a, b) on the graph. Since b = f (a), then f −1 (b) = a. Therefore, when we graph
f −1, the point (b, a) is on the graph. As a result, the graph of f −1 is a reflection of the graph of f about the line
y = x.
Figure 1.39 (a) The graph of this function f shows point (a, b) on the graph of f . (b)
Since (a, b) is on the graph of f , the point (b, a) is on the graph of f −1. The graph of
f −1 is a reflection of the graph of f about the line y = x.
Example 1.30
Sketching Graphs of Inverse Functions
For the graph of f in the following image, sketch a graph of f −1 by sketching the line y = x and using
symmetry. Identify the domain and range of f −1.
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Chapter 1 | Functions and Graphs
Solution
Reflect the graph about the line y = x. The domain of f −1 is [0, ∞). The range of f −1 is [−2, ∞). By using
the preceding strategy for finding inverse functions, we can verify that the inverse function is f −1 (x) = x 2 − 2,
as shown in the graph.
1.25
Sketch the graph of f (x) = 2x + 3 and the graph of its inverse using the symmetry property of inverse
functions.
Restricting Domains
As we have seen, f (x) = x 2 does not have an inverse function because it is not one-to-one. However, we can choose a
subset of the domain of f such that the function is one-to-one. This subset is called a restricted domain. By restricting the
domain of f , we can define a new function g such that the domain of g is the restricted domain of f and g(x) = f (x)
for all x in the domain of g. Then we can define an inverse function for g on that domain. For example, since f (x) = x 2
is one-to-one on the interval [0, ∞), we can define a new function g such that the domain of g is [0, ∞) and g(x) = x 2
for all x in its domain. Since g is a one-to-one function, it has an inverse function, given by the formula g −1(x) = x. On
the other hand, the function f (x) = x 2 is also one-to-one on the domain (−∞, 0]. Therefore, we could also define a new
function h such that the domain of h is (−∞, 0] and h(x) = x 2 for all x in the domain of h. Then h is a one-to-one
function and must also have an inverse. Its inverse is given by the formula h −1(x) = − x (Figure 1.40).
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Chapter 1 | Functions and Graphs
Figure 1.40 (a) For g(x) = x 2 restricted to [0, ∞), g −1 (x) = x. (b) For
h(x) = x 2 restricted to (−∞, 0], h −1 (x) = − x.
Example 1.31
Restricting the Domain
Consider the function f (x) = (x + 1) 2.
a. Sketch the graph of f and use the horizontal line test to show that f is not one-to-one.
b. Show that f is one-to-one on the restricted domain [−1, ∞). Determine the domain and range for the
inverse of f on this restricted domain and find a formula for f −1.
Solution
a. The graph of f is the graph of y = x 2 shifted left 1 unit. Since there exists a horizontal line intersecting
the graph more than once, f is not one-to-one.
b. On the interval [−1, ∞), f is one-to-one.
83
84
Chapter 1 | Functions and Graphs
The domain and range of f −1 are given by the range and domain of f , respectively. Therefore, the
domain of f −1 is [0, ∞) and the range of f −1 is [−1, ∞). To find a formula for f −1, solve the
equation y = (x + 1) 2 for x. If y = (x + 1) 2, then x = −1 ± y. Since we are restricting the domain
to the interval where x ≥ −1, we need ± y ≥ 0. Therefore, x = −1 + y. Interchanging x and y,
we write y = −1 + x and conclude that f −1 (x) = −1 + x.
1.26
Consider f (x) = 1/x 2 restricted to the domain (−∞, 0). Verify that f is one-to-one on this domain.
Determine the domain and range of the inverse of f and find a formula for f −1.
Inverse Trigonometric Functions
The six basic trigonometric functions are periodic, and therefore they are not one-to-one. However, if we restrict the domain
of a trigonometric function to an interval where it is one-to-one, we can define its inverse. Consider the sine function
(Figure 1.34). The sine function is one-to-one on an infinite number of intervals, but the standard convention is to restrict
⎡
⎤
the domain to the interval ⎣− π , π ⎦. By doing so, we define the inverse sine function on the domain [−1, 1] such that
2 2
⎡
⎤
for any x in the interval [−1, 1], the inverse sine function tells us which angle θ in the interval ⎣− π , π ⎦ satisfies
2 2
sinθ = x. Similarly, we can restrict the domains of the other trigonometric functions to define inverse trigonometric
functions, which are functions that tell us which angle in a certain interval has a specified trigonometric value.
Definition
The inverse sine function, denoted sin −1 or arcsin, and the inverse cosine function, denoted cos −1 or arccos, are
defined on the domain D = {x| − 1 ≤ x ≤ 1} as follows:
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Chapter 1 | Functions and Graphs
85
sin −1 (x) = y if and only if sin(y) = x and − π ≤ y ≤ π ;
2
2
cos −1 (x) = y if and only if cos(y) = x and 0 ≤ y ≤ π.
(1.12)
The inverse tangent function, denoted tan −1 or arctan, and inverse cotangent function, denoted cot −1 or arccot, are
defined on the domain D = {x| − ∞ < x < ∞} as follows:
tan −1 (x) = y if and only if tan(y) = x and − π < y < π ;
2
2
−1
cot (x) = y if and only if cot(y) = x and 0 < y < π.
(1.13)
The inverse cosecant function, denoted csc −1 or arccsc, and inverse secant function, denoted sec −1 or arcsec, are
defined on the domain D = {x||x| ≥ 1} as follows:
csc −1 (x) = y if and only if csc(y) = x and − π ≤ y ≤ π , y ≠ 0;
2
2
−1
sec (x) = y if and only if sec(y) = x and 0 ≤ y ≤ π, y ≠ π/2.
(1.14)
To graph the inverse trigonometric functions, we use the graphs of the trigonometric functions restricted to the domains
defined earlier and reflect the graphs about the line y = x (Figure 1.41).
Figure 1.41 The graph of each of the inverse trigonometric functions is a reflection about the line y = x of
the corresponding restricted trigonometric function.
Go to the following site (http://www.openstaxcollege.org/l/20_inversefun) for more comparisons of
functions and their inverses.
86
Chapter 1 | Functions and Graphs
⎛ ⎞
When evaluating an inverse trigonometric function, the output is an angle. For example, to evaluate cos −1 ⎝1 ⎠, we need to
2
1
find an angle θ such that cosθ = . Clearly, many angles have this property. However, given the definition of cos −1,
2
⎛ ⎞
need the angle θ that not only solves this equation, but also lies in the interval [0, π]. We conclude that cos −1 ⎝1 ⎠ =
2
we
π.
3
We now consider a composition of a trigonometric function and its inverse. For example, consider the two expressions
⎛ ⎞⎞
⎛
sin⎝sin −1 ⎝ 2 ⎠⎠ and sin −1(sin(π)). For the first one, we simplify as follows:
2
⎛ ⎞⎞
⎛
⎛ ⎞
sin⎝sin −1 ⎝ 2 ⎠⎠ = sin⎝π ⎠ = 2 .
2
2
4
For the second one, we have
sin −1 ⎛⎝sin(π)⎞⎠ = sin −1 (0) = 0.
The inverse function is supposed to “undo” the original function, so why isn’t sin −1 ⎛⎝sin(π)⎞⎠ = π ? Recalling our definition
of inverse functions, a function f and its inverse f −1 satisfy the conditions f ⎛⎝ f −1 (y)⎞⎠ = y for all y in the domain of
f −1 and f −1 ⎛⎝ f (x)⎞⎠ = x for all x in the domain of f , so what happened here? The issue is that the inverse sine function,
⎡
⎤
sin −1, is the inverse of the restricted sine function defined on the domain ⎣− π , π ⎦. Therefore, for x in the interval
2 2
⎡ π π⎤
−1
⎣− 2 , 2 ⎦, it is true that sin (sin x) = x. However, for values of x outside this interval, the equation does not hold, even
though sin −1(sin x) is defined for all real numbers x.
What about sin(sin −1 y) ? Does that have a similar issue? The answer is no. Since the domain of sin −1 is the interval
[−1, 1], we conclude that sin(sin −1 y) = y if −1 ≤ y ≤ 1 and the expression is not defined for other values of y. To
summarize,
sin(sin −1 y) = y if −1 ≤ y ≤ 1
and
sin −1 (sin x) = x if − π ≤ x ≤ π .
2
2
Similarly, for the cosine function,
cos(cos −1 y) = y if −1 ≤ y ≤ 1
and
cos −1 (cos x) = x if 0 ≤ x ≤ π.
Similar properties hold for the other trigonometric functions and their inverses.
Example 1.32
Evaluating Expressions Involving Inverse Trigonometric Functions
Evaluate each of the following expressions.
a.
b.
⎛
⎞
sin −1 ⎝− 3 ⎠
2
⎛
⎞⎞
⎛
tan⎝tan −1 ⎝− 1 ⎠⎠
3
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Chapter 1 | Functions and Graphs
87
⎛
⎛
⎞⎞
⎛
⎛
⎞⎞
c.
cos −1 ⎝cos⎝5π ⎠⎠
4
d.
sin −1 ⎝cos⎝2π ⎠⎠
3
Solution
a. Evaluating sin −1 ⎛⎝− 3/2⎞⎠ is equivalent to finding the angle θ such that sinθ = − 3/2 and
−π/2 ≤ θ ≤ π/2.
The
angle
sin −1 ⎛⎝− 3/2⎞⎠ = −π/3.
b. First
we
use
tan⎛⎝tan −1 ⎛⎝−1/
the
⎞⎞
⎠⎠
fact
that
θ = −π/3
satisfies
these
two
conditions.
Therefore,
tan −1 ⎛⎝−1/ 3⎞⎠ = −π/6.
Then
tan(π/6) = −1/ 3.
Therefore,
3 = −1/ 3.
c. To evaluate cos −1 ⎛⎝cos(5π/4)⎞⎠, first use the fact that cos(5π/4) = − 2/2. Then we need to find the
angle θ such that cos(θ) = − 2/2 and 0 ≤ θ ≤ π. Since 3π/4 satisfies both these conditions, we have
cos⎛⎝cos −1 (5π/4)⎞⎠ = cos⎛⎝cos −1 ⎛⎝− 2/2⎞⎠⎞⎠ = 3π/4.
d. Since cos(2π/3) = −1/2, we need to evaluate sin −1 (−1/2). That is, we need to find the angle θ such
that sin(θ) = −1/2 and −π/2 ≤ θ ≤ π/2. Since −π/6 satisfies both these conditions, we can conclude
that sin −1 (cos(2π/3)) = sin −1 (−1/2) = −π/6.
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Chapter 1 | Functions and Graphs
The Maximum Value of a Function
In many areas of science, engineering, and mathematics, it is useful to know the maximum value a function can obtain,
even if we don’t know its exact value at a given instant. For instance, if we have a function describing the strength
of a roof beam, we would want to know the maximum weight the beam can support without breaking. If we have a
function that describes the speed of a train, we would want to know its maximum speed before it jumps off the rails.
Safe design often depends on knowing maximum values.
This project describes a simple example of a function with a maximum value that depends on two equation coefficients.
We will see that maximum values can depend on several factors other than the independent variable x.
1. Consider the graph in Figure 1.42 of the function y = sin x + cos x. Describe its overall shape. Is it periodic?
How do you know?
Figure 1.42 The graph of y = sin x + cos x.
Using a graphing calculator or other graphing device, estimate the x - and y -values of the maximum point for
the graph (the first such point where x > 0). It may be helpful to express the x -value as a multiple of π.
2. Now consider other graphs of the form y = Asin x + Bcos x for various values of A and B. Sketch the graph
when A = 2 and B = 1, and find the x - and y-values for the maximum point. (Remember to express the x-value
as a multiple of π, if possible.) Has it moved?
3. Repeat for A = 1, B = 2. Is there any relationship to what you found in part (2)?
4. Complete the following table, adding a few choices of your own for A and B:
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Chapter 1 | Functions and Graphs
89
A
B
x
y
A
0
1
1
0
1
1
1
12
5
1
2
5
12
2
1
2
2
3
4
4
3
3
B
x
y
1
3
5. Try to figure out the formula for the y-values.
6. The formula for the
⎛
⎝
⎞ ⎛
⎠ ⎝
(1, 1), 1, 3 ,
⎞
⎠
x -values is a little harder. The most helpful points from the table are
3, 1 . (Hint: Consider inverse trigonometric functions.)
7. If you found formulas for parts (5) and (6), show that they work together. That is, substitute the x -value
formula you found into y = Asin x + Bcos x and simplify it to arrive at the y -value formula you found.
90
Chapter 1 | Functions and Graphs
1.4 EXERCISES
For the following exercises, use the horizontal line test to
determine whether each of the given graphs is one-to-one.
186.
183.
187.
184.
188.
185.
For the following exercises, a. find the inverse function,
and b. find the domain and range of the inverse function.
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189.
f (x) = x 2 − 4, x ≥ 0
190.
f (x) = x − 4
191.
f (x) = x 3 + 1
192.
f (x) = (x − 1) 2, x ≤ 1
3
Chapter 1 | Functions and Graphs
193.
f (x) = x − 1
194.
f (x) =
91
198.
1
x+2
For the following exercises, use the graph of f to sketch
the graph of its inverse function.
195.
For the following exercises, use composition to determine
which pairs of functions are inverses.
196.
199.
f (x) = 8x, g(x) = x
8
200.
f (x) = 8x + 3, g(x) = x − 3
8
201.
f (x) = 5x − 7, g(x) = x + 5
7
202.
f (x) = 2 x + 2, g(x) = 3 x + 3
3
2
203.
f (x) =
204.
f (x) = x 3 + 1, g(x) = (x − 1) 1/3
1 , x ≠ 1, g(x) = 1 + 1, x ≠ 0
x
x−1
205.
f (x) = x 2 + 2x + 1, x ≥ −1,
197.
g(x) = −1 + x, x ≥ 0
206.
f (x) = 4 − x 2, 0 ≤ x ≤ 2, g(x) = 4 − x 2, 0 ≤ x ≤ 2
For the following exercises, evaluate the functions. Give
the exact value.
⎛ ⎞
207. tan −1 ⎝ 3 ⎠
3
⎛
⎞
208. cos −1 ⎝− 2 ⎠
2
209. cot −1 (1)
210. sin −1 (−1)
⎛ ⎞
211. cos −1 ⎝ 3 ⎠
2
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Chapter 1 | Functions and Graphs
212. cos⎛⎝tan −1 ⎛⎝ 3⎞⎠⎞⎠
⎛ ⎞⎞
⎛
213. sin⎝cos −1 ⎝ 2 ⎠⎠
2
⎛
⎛ ⎞⎞
⎛
⎛
221. [T] An airplane’s Mach number M is the ratio of
its speed to the speed of sound. When a plane is flying
at a constant altitude, then its Mach angle is given by
⎛
⎞
µ = 2sin −1 ⎝ 1 ⎠. Find the Mach angle (to the nearest
M
degree) for the following Mach numbers.
214. sin −1 ⎝sin⎝π ⎠⎠
3
⎞⎞
215. tan −1 ⎝tan⎝− π ⎠⎠
6
216. The function C = T(F) = (5/9)(F − 32) converts
degrees Fahrenheit to degrees Celsius.
a. Find the inverse function F = T −1(C)
b. What is the inverse function used for?
217. [T] The velocity V (in centimeters per second) of
blood in an artery at a distance x cm from the center of
the artery can be modeled by the function
V = f (x) = 500(0.04 − x 2) for 0 ≤ x ≤ 0.2.
a. Find x = f −1(V).
b. Interpret what the inverse function is used for.
c. Find the distance from the center of an artery with
a velocity of 15 cm/sec, 10 cm/sec, and 5 cm/sec.
218. A function that converts dress sizes in the United
States to those in Europe is given by D(x) = 2x + 24.
a. Find the European dress sizes that correspond to
sizes 6, 8, 10, and 12 in the United States.
b. Find the function that converts European dress
sizes to U.S. dress sizes.
c. Use part b. to find the dress sizes in the United
States that correspond to 46, 52, 62, and 70.
219. [T] The cost to remove a toxin from a lake is
modeled by the function C(p) = 75p/(85 − p), where
C is the cost (in thousands of dollars) and p is the amount
of toxin in a small lake (measured in parts per billion
[ppb]). This model is valid only when the amount of toxin
is less than 85 ppb.
a. Find the cost to remove 25 ppb, 40 ppb, and 50 ppb
of the toxin from the lake.
b. Find the inverse function. c. Use part b. to
determine how much of the toxin is removed for
$50,000.
220. [T] A race car is accelerating at a velocity given
by v(t) = 25 t + 54, where v is the velocity (in feet per
4
second) at time t.
a. Find the velocity of the car at 10 sec.
b. Find the inverse function.
c. Use part b. to determine how long it takes for the
car to reach a speed of 150 ft/sec.
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a.
µ = 1.4
b.
µ = 2.8
c.
µ = 4.3
⎛
⎞
222. [T] Using µ = 2sin −1 ⎝ 1 ⎠, find the Mach number
M
M for the following angles.
a. µ = π
b.
c.
6
2π
µ=
7
3π
µ=
8
223. [T] The temperature (in degrees Celsius) of a city in
the northern United States can be modeled by the function
⎡
⎤
T(x) = 5 + 18sin⎣π (x − 4.6)⎦, where x is time in
6
months and x = 1.00 corresponds to January 1. Determine
the month and day when the temperature is 21°C.
224. [T] The depth (in feet) of water at a dock changes
with the rise and fall of tides. It is modeled by the function
⎛
⎞
D(t) = 5sin⎝π t − 7π ⎠ + 8, where t is the number of
6
6
hours after midnight. Determine the first time after
midnight when the depth is 11.75 ft.
225. [T] An object moving in simple harmonic motion
⎛
⎞
is modeled by the function s(t) = −6cos⎝πt ⎠, where s
2
is measured in inches and t is measured in seconds.
Determine the first time when the distance moved is 4.5 ft.
Chapter 1 | Functions and Graphs
226. [T] A local art gallery has a portrait 3 ft in height
that is hung 2.5 ft above the eye level of an average person.
The viewing angle θ can be modeled by the function
−1 2.5
θ = tan −1 5.5
x − tan
x , where x is the distance (in
feet) from the portrait. Find the viewing angle when a
person is 4 ft from the portrait.
227. [T] Use a calculator to evaluate tan −1 (tan(2.1)) and
cos −1 (cos(2.1)). Explain the results of each.
228. [T] Use a calculator to evaluate sin(sin −1(−2)) and
tan(tan −1(−2)). Explain the results of each.
93
94
Chapter 1 | Functions and Graphs
1.5 | Exponential and Logarithmic Functions
Learning Objectives
1.5.1 Identify the form of an exponential function.
1.5.2 Explain the difference between the graphs of x b and b x.
1.5.3 Recognize the significance of the number e.
1.5.4
1.5.5
1.5.6
1.5.7
Identify the form of a logarithmic function.
Explain the relationship between exponential and logarithmic functions.
Describe how to calculate a logarithm to a different base.
Identify the hyperbolic functions, their graphs, and basic identities.
In this section we examine exponential and logarithmic functions. We use the properties of these functions to solve
equations involving exponential or logarithmic terms, and we study the meaning and importance of the number e. We also
define hyperbolic and inverse hyperbolic functions, which involve combinations of exponential and logarithmic functions.
(Note that we present alternative definitions of exponential and logarithmic functions in the chapter Applications of
Integrations, and prove that the functions have the same properties with either definition.)
Exponential Functions
Exponential functions arise in many applications. One common example is population growth.
For example, if a population starts with P 0 individuals and then grows at an annual rate of 2%, its population after 1 year
is
P(1) = P 0 + 0.02P 0 = P 0(1 + 0.02) = P 0(1.02).
Its population after 2 years is
P(2) = P(1) + 0.02P(1) = P(1)(1.02) = P 0 (1.02) 2.
In general, its population after t years is
P(t) = P 0 (1.02) t,
which is an exponential function. More generally, any function of the form f (x) = b x, where b > 0, b ≠ 1, is an
exponential function with base b and exponent x. Exponential functions have constant bases and variable exponents. Note
that a function of the form f (x) = x b for some constant b is not an exponential function but a power function.
To see the difference between an exponential function and a power function, we compare the functions y = x 2 and y = 2 x.
In Table 1.10, we see that both 2 x and x 2 approach infinity as x → ∞. Eventually, however, 2 x becomes larger than
x 2 and grows more rapidly as x → ∞. In the opposite direction, as x → −∞, x 2 → ∞, whereas 2 x → 0. The line
y = 0 is a horizontal asymptote for y = 2 x.
x
−3
−2
−1
0
1
2
3
4
5
6
x2
9
4
1
0
1
4
9
16
25
36
2x
1/8
1/4
1/2
1
2
4
8
16
32
64
Table 1.10 Values of x 2 and 2 x
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Chapter 1 | Functions and Graphs
95
In Figure 1.43, we graph both y = x 2 and y = 2 x to show how the graphs differ.
Figure 1.43 Both 2 x and x 2 approach infinity as x → ∞,
but 2 x grows more rapidly than x 2. As
x → −∞, x 2 → ∞, whereas 2 x → 0.
Evaluating Exponential Functions
Recall the properties of exponents: If x is a positive integer, then we define b x = b · b ⋯ b (with x factors of b). If x
is a negative integer, then x = −y for some positive integer y, and we define b x = b
−y
= 1/b y. Also, b 0 is defined
to be 1. If x is a rational number, then x = p/q, where p and q are integers and b x = b
p/q
q
= b p. For example,
9 3/2 = 9 3 = 27. However, how is b x defined if x is an irrational number? For example, what do we mean by 2 2 ?
This is too complex a question for us to answer fully right now; however, we can make an approximation. In Table 1.11,
we list some rational numbers approaching 2, and the values of 2 x for each rational number x are presented as well.
We claim that if we choose rational numbers x getting closer and closer to
2, the values of 2 x get closer and closer to
some number L. We define that number L to be 2 2.
x
1.4
1.41
1.414
1.4142
1.41421
1.414213
2x
2.639
2.65737
2.66475
2.665119
2.665138
2.665143
Table 1.11 Values of 2 x for a List of Rational Numbers Approximating
2
Example 1.33
Bacterial Growth
Suppose a particular population of bacteria is known to double in size every 4 hours. If a culture starts with
1000 bacteria, the number of bacteria after 4 hours is n(4) = 1000 · 2. The number of bacteria after 8 hours is
n(8) = n(4) · 2 = 1000 · 2 2. In general, the number of bacteria after 4m hours is n(4m) = 1000 · 2 m. Letting
96
Chapter 1 | Functions and Graphs
t = 4m, we see that the number of bacteria after t hours is n(t) = 1000 · 2 t/4. Find the number of bacteria
after 6 hours, 10 hours, and 24 hours.
Solution
The number of bacteria after 6 hours is given by n(6) = 1000 · 2 6/4 ≈ 2828 bacteria. The number of bacteria
after 10 hours is given by n(10) = 1000 · 2 10/4 ≈ 5657 bacteria. The number of bacteria after 24 hours is
given by n(24) = 1000 · 2 6 = 64,000 bacteria.
1.27
Given the exponential function f (x) = 100 · 3 x/2, evaluate f (4) and f (10).
Go to World Population Balance (http://www.openstaxcollege.org/l/20_exponengrow) for another
example of exponential population growth.
Graphing Exponential Functions
For any base b > 0, b ≠ 1, the exponential function f (x) = b x is defined for all real numbers x and b x > 0. Therefore,
the domain of f (x) = b x is (−∞, ∞) and the range is (0, ∞). To graph b x, we note that for b > 1, b x is increasing
on (−∞, ∞) and b x → ∞ as x → ∞, whereas b x → 0 as x → −∞. On the other hand, if 0 < b < 1, f (x) = b x is
decreasing on (−∞, ∞) and b x → 0 as x → ∞ whereas b x → ∞ as x → −∞ (Figure 1.44).
Figure 1.44 If b > 1, then b x is increasing on (−∞, ∞).
If 0 < b < 1, then b x is decreasing on (−∞, ∞).
Visit this site (http://www.openstaxcollege.org/l/20_inverse) for more exploration of the graphs of
exponential functions.
Note that exponential functions satisfy the general laws of exponents. To remind you of these laws, we state them as rules.
Rule: Laws of Exponents
For any constants a > 0, b > 0, and for all x and y,
x+y
1.
bx · by = b
2.
bx = bx − y
by
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Chapter 1 | Functions and Graphs
97
xy
3.
(b x) y = b
4.
(ab) x = a x b x
5.
a x = ⎛a ⎞
b x ⎝b ⎠
x
Example 1.34
Using the Laws of Exponents
Use the laws of exponents to simplify each of the following expressions.
a.
b.
⎛ 2/3⎞
⎠
⎝2x
3
⎛ −1/3⎞
⎠
⎝4x
⎛ 3 −1⎞
⎝x y ⎠
⎛ 2⎞
⎝xy ⎠
2
2
−2
Solution
a. We can simplify as follows:
⎛ 2/3⎞
⎠
⎝2x
3
2
⎞
⎛ −1/3
⎠
⎝4x
=
2 3 ⎛⎝x 2/3⎞⎠
3
2
⎞
4 2 ⎛⎝x −1/3⎠
=
8x 2 = x 2 x 2/3 = x 8/3 .
2
2
16x −2/3
b. We can simplify as follows:
⎛ 3 −1⎞
⎝x y ⎠
⎛ 2⎞
⎝xy ⎠
1.28
2
−2
2
=
⎛ 3⎞ ⎛ −1⎞
⎝x ⎠ ⎝y ⎠
x −2 ⎛⎝y 2⎞⎠
2
−2
=
x 6 y −2
= x 6 x 2 y −2 y 4 = x 8 y 2.
x −2 y −4
Use the laws of exponents to simplify ⎛⎝6x −3 y 2⎞⎠/⎛⎝12x −4 y 5⎞⎠.
The Number e
A special type of exponential function appears frequently in real-world applications. To describe it, consider the following
example of exponential growth, which arises from compounding interest in a savings account. Suppose a person invests P
dollars in a savings account with an annual interest rate r, compounded annually. The amount of money after 1 year is
A(1) = P + rP = P(1 + r).
The amount of money after 2 years is
A(2) = A(1) + rA(1) = P(1 + r) + rP(1 + r) = P(1 + r) 2.
More generally, the amount after t years is
A(t) = P(1 + r) t.
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Chapter 1 | Functions and Graphs
If the money is compounded 2 times per year, the amount of money after half a year is
⎛ ⎞
⎛ ⎞
⎛
⎛ ⎞⎞
A⎝1 ⎠ = P + ⎝ r ⎠P = P⎝1 + ⎝ r ⎠⎠.
2
2
2
The amount of money after 1 year is
⎞
⎛ ⎛
⎞⎞
⎞
⎛ ⎞ ⎛ ⎞ ⎛ ⎞
⎛
⎛
A(1) = A⎝1 ⎠ + ⎝ r ⎠A⎝1 ⎠ = P⎝1 + r ⎠ + r ⎝P⎝1 + r ⎠⎠ = P⎝1 + r ⎠ .
2
2
2 2
2
2
2
2
After t years, the amount of money in the account is
⎛
⎞
A(t) = P⎝1 + r ⎠ .
2
2t
More generally, if the money is compounded n times per year, the amount of money in the account after t years is given
by the function
nt
A(t) = P⎛⎝1 + nr ⎞⎠ .
What happens as n → ∞ ? To answer this question, we let m = n/r and write
nt
1⎞
1 + nr ⎞⎠ = ⎛⎝1 + m
⎠
⎛
⎝
mrt
,
and examine the behavior of (1 + 1/m) m as m → ∞, using a table of values (Table 1.12).
m
⎛
⎝1 +
m
1⎞
m⎠
10
100
1000
10,000
100,000
1,000,000
2.5937
2.7048
2.71692
2.71815
2.718268
2.718280
⎛
1⎞
Table 1.12 Values of ⎝1 + m
⎠
m
as m → ∞
Looking at this table, it appears that (1 + 1/m) m is approaching a number between 2.7 and 2.8 as m → ∞. In fact,
(1 + 1/m) m does approach some number as m → ∞. We call this number e . To six decimal places of accuracy,
e ≈ 2.718282.
The letter e was first used to represent this number by the Swiss mathematician Leonhard Euler during the 1720s. Although
Euler did not discover the number, he showed many important connections between e and logarithmic functions. We still
use the notation e today to honor Euler’s work because it appears in many areas of mathematics and because we can use it
in many practical applications.
Returning to our savings account example, we can conclude that if a person puts P dollars in an account at an annual
interest rate r, compounded continuously, then A(t) = Pe rt. This function may be familiar. Since functions involving
base e arise often in applications, we call the function f (x) = e x the natural exponential function. Not only is this
function interesting because of the definition of the number e, but also, as discussed next, its graph has an important
property.
Since e > 1, we know e x is increasing on (−∞, ∞). In Figure 1.45, we show a graph of f (x) = e x along with a
tangent line to the graph of at x = 0. We give a precise definition of tangent line in the next chapter; but, informally, we
say a tangent line to a graph of f at x = a is a line that passes through the point ⎛⎝a, f (a)⎞⎠ and has the same “slope” as
f at that point . The function f (x) = e x is the only exponential function b x with tangent line at x = 0 that has a slope
of 1. As we see later in the text, having this property makes the natural exponential function the most simple exponential
function to use in many instances.
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99
Figure 1.45 The graph of f (x) = e x has a tangent line with
slope 1 at x = 0.
Example 1.35
Compounding Interest
Suppose $500 is invested in an account at an annual interest rate of r = 5.5%, compounded continuously.
a. Let t denote the number of years after the initial investment and A(t) denote the amount of money in
the account at time t. Find a formula for A(t).
b. Find the amount of money in the account after 10 years and after 20 years.
Solution
a. If P dollars are invested in an account at an annual interest rate r, compounded continuously, then
A(t) = Pe rt. Here P = $500 and r = 0.055. Therefore, A(t) = 500e 0.055t.
b. After 10 years, the amount of money in the account is
A(10) = 500e 0.055 · 10 = 500e 0.55 ≈ $866.63.
After 20 years, the amount of money in the account is
A(20) = 500e 0.055 · 20 = 500e 1.1 ≈ $1, 502.08.
1.29
If $750 is invested in an account at an annual interest rate of 4%, compounded continuously, find a
formula for the amount of money in the account after t years. Find the amount of money after 30 years.
Logarithmic Functions
Using our understanding of exponential functions, we can discuss their inverses, which are the logarithmic functions. These
come in handy when we need to consider any phenomenon that varies over a wide range of values, such as pH in chemistry
or decibels in sound levels.
The exponential function f (x) = b x is one-to-one, with domain (−∞, ∞) and range (0, ∞). Therefore, it has an inverse
function, called the logarithmic function with base b. For any b > 0, b ≠ 1, the logarithmic function with base b,
denoted log b, has domain (0, ∞) and range (−∞, ∞), and satisfies
log b (x) = y if and only if b y = x.
For example,
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Chapter 1 | Functions and Graphs
⎛
⎞
log 10 ⎝ 1 ⎠ = −2
100
since
2 3 = 8,
since
log b(1) = 0
since
10 −2 = 1 2 = 1 ,
100
10
0
b = 1 for any base b > 0.
log 2(8) = 3
Furthermore, since y = log b(x) and y = b x are inverse functions,
log b (b x) = x and b
log b (x)
= x.
The most commonly used logarithmic function is the function log e. Since this function uses natural e as its base, it is
called the natural logarithm. Here we use the notation ln(x) or ln x to mean log e (x). For example,
ln(e) = log e (e) = 1, ln⎛⎝e 3⎞⎠ = log e ⎛⎝e 3⎞⎠ = 3, ln(1) = log e (1) = 0.
Since the functions f (x) = e x and g(x) = ln(x) are inverses of each other,
ln(e x) = x and e ln x = x,
and their graphs are symmetric about the line y = x (Figure 1.46).
Figure 1.46 The functions y = e x and y = ln(x) are
inverses of each other, so their graphs are symmetric about the
line y = x.
At this site (http://www.openstaxcollege.org/l/20_logscale) you can see an example of a base-10
logarithmic scale.
In general, for any base b > 0, b ≠ 1, the function g(x) = log b(x) is symmetric about the line y = x with the function
f (x) = b x. Using this fact and the graphs of the exponential functions, we graph functions log b for several values of
b > 1 (Figure 1.47).
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101
Figure 1.47 Graphs of y = log b(x) are depicted for
b = 2, e, 10.
Before solving some equations involving exponential and logarithmic functions, let’s review the basic properties of
logarithms.
Rule: Properties of Logarithms
If a, b, c > 0, b ≠ 1, and r is any real number, then
1.
2.
3.
log b (ac) = log b (a) + log b (c)
log b ⎛⎝ac ⎞⎠ = log b (a) − log b (c)
log b (a r) = rlog b (a)
(Product property)
(Quotient property)
(Power property)
Example 1.36
Solving Equations Involving Exponential Functions
Solve each of the following equations for x.
a.
5x = 2
b.
e x + 6e −x = 5
Solution
a. Applying the natural logarithm function to both sides of the equation, we have
ln5 x = ln2.
Using the power property of logarithms,
x ln5 = ln2.
Therefore, x = ln2/ln5.
b. Multiplying both sides of the equation by e x, we arrive at the equation
e 2x + 6 = 5e x.
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Chapter 1 | Functions and Graphs
Rewriting this equation as
e 2x − 5e x + 6 = 0,
we can then rewrite it as a quadratic equation in e x :
(e x) 2 − 5(e x) + 6 = 0.
Now we can solve the quadratic equation. Factoring this equation, we obtain
(e x − 3)(e x − 2) = 0.
Therefore, the solutions satisfy e x = 3 and e x = 2. Taking the natural logarithm of both sides gives us
the solutions x = ln3, ln2.
1.30
Solve e 2x /(3 + e 2x) = 1/2.
Example 1.37
Solving Equations Involving Logarithmic Functions
Solve each of the following equations for x.
a.
ln⎛⎝1x ⎞⎠ = 4
b.
log 10 x + log 10 x = 2
c.
ln(2x) − 3ln⎛⎝x 2⎞⎠ = 0
Solution
a. By the definition of the natural logarithm function,
ln⎛⎝1x ⎞⎠ = 4 if and only if e 4 = 1x .
Therefore, the solution is x = 1/e 4.
b. Using the product and power properties of logarithmic functions, rewrite the left-hand side of the equation
as
log 10 x + log 10 x = log 10 x x = log 10 x 3/2 = 3 log 10 x.
2
Therefore, the equation can be rewritten as
3 log x = 2 or log x = 4 .
10
2 10
3
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103
3
The solution is x = 10 4/3 = 10 10.
⎛
⎞
c. Using the power property of logarithmic functions, we can rewrite the equation as ln(2x) − ln⎝x 6⎠ = 0.
Using the quotient property, this becomes
⎛ ⎞
ln 25 = 0.
⎝x ⎠
5
Therefore, 2/x 5 = 1, which implies x = 2. We should then check for any extraneous solutions.
1.31
Solve ln⎛⎝x 3⎞⎠ − 4ln(x) = 1.
When evaluating a logarithmic function with a calculator, you may have noticed that the only options are log 10 or log,
called the common logarithm, or ln, which is the natural logarithm. However, exponential functions and logarithm functions
can be expressed in terms of any desired base b. If you need to use a calculator to evaluate an expression with a different
base, you can apply the change-of-base formulas first. Using this change of base, we typically write a given exponential or
logarithmic function in terms of the natural exponential and natural logarithmic functions.
Rule: Change-of-Base Formulas
Let a > 0, b > 0, and a ≠ 1, b ≠ 1.
1.
ax = b
xlog b a
for any real number x.
If b = e, this equation reduces to a x = e
2.
log a x =
xlog e a
= e x lna.
log b x
for any real number x > 0.
log b a
If b = e, this equation reduces to log a x = ln x .
lna
Proof
For the first change-of-base formula, we begin by making use of the power property of logarithmic functions. We know that
for any base b > 0, b ≠ 1, log b(a x) = xlog b a. Therefore,
b
log b(a x)
=b
xlog b a
.
x
In addition, we know that b and log b(x) are inverse functions. Therefore,
b
log b(a x)
Combining these last two equalities, we conclude that a x = b
= a x.
xlog b a
.
To prove the second property, we show that
(log b a) · (log a x) = log b x.
Let u = log b a, v = log a x, and w = log b x. We will show that u · v = w. By the definition of logarithmic functions, we
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Chapter 1 | Functions and Graphs
know that b u = a, a v = x, and b w = x. From the previous equations, we see that
b uv = (b u) v = a v = x = b w.
Therefore, b uv = b w. Since exponential functions are one-to-one, we can conclude that u · v = w.
□
Example 1.38
Changing Bases
Use a calculating utility to evaluate log 3 7 with the change-of-base formula presented earlier.
Solution
Use the second equation with a = 3 and e = 3:
log 3 7 = ln7 ≈ 1.77124.
ln3
1.32
Use the change-of-base formula and a calculating utility to evaluate log 4 6.
Example 1.39
Chapter Opener: The Richter Scale for Earthquakes
Figure 1.48 (credit: modification of work by Robb
Hannawacker, NPS)
In 1935, Charles Richter developed a scale (now known as the Richter scale) to measure the magnitude of an
earthquake. The scale is a base-10 logarithmic scale, and it can be described as follows: Consider one earthquake
with magnitude R 1 on the Richter scale and a second earthquake with magnitude R 2 on the Richter scale.
Suppose R 1 > R 2, which means the earthquake of magnitude R 1 is stronger, but how much stronger is it than
the other earthquake? A way of measuring the intensity of an earthquake is by using a seismograph to measure
the amplitude of the earthquake waves. If A 1 is the amplitude measured for the first earthquake and A 2 is the
amplitude measured for the second earthquake, then the amplitudes and magnitudes of the two earthquakes satisfy
the following equation:
R 1 − R 2 = log 10
⎛A1 ⎞
⎝ A 2 ⎠.
Consider an earthquake that measures 8 on the Richter scale and an earthquake that measures 7 on the Richter
scale. Then,
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105
8 − 7 = log 10
Therefore,
log 10
⎛A1 ⎞
⎝ A 2 ⎠.
⎛A1 ⎞
⎝ A 2 ⎠ = 1,
which implies A 1 /A 2 = 10 or A 1 = 10A 2. Since A 1 is 10 times the size of A 2, we say that the first
earthquake is 10 times as intense as the second earthquake. On the other hand, if one earthquake measures 8 on
the Richter scale and another measures 6, then the relative intensity of the two earthquakes satisfies the equation
log 10
⎛A1 ⎞
⎝ A 2 ⎠ = 8 − 6 = 2.
Therefore, A 1 = 100A 2. That is, the first earthquake is 100 times more intense than the second earthquake.
How can we use logarithmic functions to compare the relative severity of the magnitude 9 earthquake in Japan in
2011 with the magnitude 7.3 earthquake in Haiti in 2010?
Solution
To compare the Japan and Haiti earthquakes, we can use an equation presented earlier:
9 − 7.3 = log 10
⎛A1 ⎞
⎝ A 2 ⎠.
Therefore, A 1 /A 2 = 10 1.7, and we conclude that the earthquake in Japan was approximately 50 times more
intense than the earthquake in Haiti.
1.33
Compare the relative severity of a magnitude 8.4 earthquake with a magnitude 7.4 earthquake.
Hyperbolic Functions
The hyperbolic functions are defined in terms of certain combinations of e x and e −x. These functions arise naturally
in various engineering and physics applications, including the study of water waves and vibrations of elastic membranes.
Another common use for a hyperbolic function is the representation of a hanging chain or cable, also known as a catenary
(Figure 1.49). If we introduce a coordinate system so that the low point of the chain lies along the y -axis, we can describe
the height of the chain in terms of a hyperbolic function. First, we define the hyperbolic functions.
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Chapter 1 | Functions and Graphs
Figure 1.49 The shape of a strand of silk in a spider’s web
can be described in terms of a hyperbolic function. The same
shape applies to a chain or cable hanging from two supports with
only its own weight. (credit: “Mtpaley”, Wikimedia Commons)
Definition
Hyperbolic cosine
−x
x
cosh x = e + e
2
Hyperbolic sine
−x
x
sinh x = e − e
2
Hyperbolic tangent
−x
x
tanh x = sinh x = e x − e −x
cosh x e + e
Hyperbolic cosecant
csch x =
1 =
2
sinh x e x − e −x
sech x =
1 =
2
cosh x e x + e −x
Hyperbolic secant
Hyperbolic cotangent
−x
x
coth x = cosh x = e x + e −x
sinh x e − e
The name cosh rhymes with “gosh,” whereas the name sinh is pronounced “cinch.” Tanh, sech, csch, and coth are
pronounced “tanch,” “seech,” “coseech,” and “cotanch,” respectively.
Using the definition of cosh(x) and principles of physics, it can be shown that the height of a hanging chain, such as the
one in Figure 1.49, can be described by the function h(x) = acosh(x/a) + c for certain constants a and c.
But why are these functions called hyperbolic functions? To answer this question, consider the quantity cosh 2 t − sinh 2 t.
Using the definition of cosh and sinh, we see that
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107
2t
−2t
2t
−2t
cosh 2 t − sinh 2 t = e + 2 + e
− e −2+e
= 1.
4
4
This identity is the analog of the trigonometric identity cos 2 t + sin 2 t = 1. Here, given a value t,
the point
(x, y) = (cosht, sinht) lies on the unit hyperbola x 2 − y 2 = 1 (Figure 1.50).
Figure 1.50 The unit hyperbola cosh 2 t − sinh 2 t = 1.
Graphs of Hyperbolic Functions
To graph cosh x and sinh x, we make use of the fact that both functions approach (1/2)e x as x → ∞, since e −x → 0
as x → ∞. As x → −∞, cosh x approaches 1/2e −x, whereas sinh x approaches −1/2e −x. Therefore, using the graphs
of 1/2e x, 1/2e −x,
and −1/2e −x as guides, we graph cosh x and sinh x. To graph tanh x,
we use the fact that
tanh(0) = 1, −1 < tanh(x) < 1 for all x, tanh x → 1 as x → ∞, and tanh x → −1 as x → −∞. The graphs of the
other three hyperbolic functions can be sketched using the graphs of cosh x, sinh x, and tanh x (Figure 1.51).
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Chapter 1 | Functions and Graphs
Figure 1.51 The hyperbolic functions involve combinations of e x and
e −x.
Identities Involving Hyperbolic Functions
The identity cosh 2 t − sinh 2 t, shown in Figure 1.50, is one of several identities involving the hyperbolic functions,
some of which are listed next. The first four properties follow easily from the definitions of hyperbolic sine and hyperbolic
cosine. Except for some differences in signs, most of these properties are analogous to identities for trigonometric functions.
Rule: Identities Involving Hyperbolic Functions
1.
cosh(−x) = cosh x
2.
sinh(−x) = −sinh x
3.
cosh x + sinh x = e x
4.
cosh x − sinh x = e −x
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109
5.
cosh 2 x − sinh 2 x = 1
6.
1 − tanh 2 x = sech 2 x
7.
coth 2 x − 1 = csch 2 x
8.
sinh(x ± y) = sinh x cosh y ± cosh x sinh y
9.
cosh(x ± y) = cosh x cosh y ± sinh x sinh y
Example 1.40
Evaluating Hyperbolic Functions
a. Simplify sinh(5ln x).
b. If sinh x = 3/4, find the values of the remaining five hyperbolic functions.
Solution
a. Using the definition of the sinh function, we write
5 ln x
− e −5ln x = e
sinh(5ln x) = e
2
⎛
⎛
⎞
ln⎝x 5⎠
b. Using the identity cosh 2 x − sinh 2 x = 1, we see that
−e
2
⎞
ln⎝x −5⎠
5
−5
= x −x .
2
⎛ ⎞
cosh 2 x = 1 + ⎝3 ⎠ = 25 .
4
16
2
Since cosh x ≥ 1 for all x, we must have cosh x = 5/4. Then, using the definitions for the other
hyperbolic functions, we conclude that tanh x = 3/5, csch x = 4/3, sech x = 4/5, and coth x = 5/3.
1.34
Simplify cosh(2ln x).
Inverse Hyperbolic Functions
From the graphs of the hyperbolic functions, we see that all of them are one-to-one except cosh x and sech x. If we
restrict the domains of these two functions to the interval [0, ∞), then all the hyperbolic functions are one-to-one, and we
can define the inverse hyperbolic functions. Since the hyperbolic functions themselves involve exponential functions, the
inverse hyperbolic functions involve logarithmic functions.
Definition
Inverse Hyperbolic Functions
⎛
⎞
sinh −1 x = arcsinh x = ln⎝x + x 2 + 1⎠
⎞
⎛
tanh −1 x = arctanh x = 1 ln⎝1 + x ⎠
2 1−x
⎛
⎞
2
sech −1 x = arcsech x = ln⎜1 + 1x − x ⎟
⎝
⎠
⎛
⎞
cosh −1 x = arccosh x = ln⎝x + x 2 − 1⎠
⎞
⎛
coth −1 x = arccot x = 1 ln⎝ x + 1 ⎠
2 x−1
⎛
⎞
2
csch −1 x = arccsch x = ln⎜1x + 1 + x ⎟
| x| ⎠
⎝
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Chapter 1 | Functions and Graphs
Let’s look at how to derive the first equation. The others follow similarly. Suppose y = sinh −1 x. Then, x = sinh y and,
y
by the definition of the hyperbolic sine function, x = e − e
−y
2
. Therefore,
e y − 2x − e
−y
= 0.
y
Multiplying this equation by e , we obtain
e
2y
− 2xe y − 1 = 0.
This can be solved like a quadratic equation, with the solution
2
e y = 2x ± 4x + 4 = x ± x 2 + 1.
2
Since e y > 0, the only solution is the one with the positive sign. Applying the natural logarithm to both sides of the
equation, we conclude that
⎛
⎞
y = ln⎝x + x 2 + 1⎠.
Example 1.41
Evaluating Inverse Hyperbolic Functions
Evaluate each of the following expressions.
sinh −1 (2)
tanh −1(1/4)
Solution
⎛
⎞
sinh −1 (2) = ln⎝2 + 2 2 + 1⎠ = ln⎛⎝2 + 5⎞⎠ ≈ 1.4436
⎛
⎛
⎛ ⎞
⎞
⎞
tanh −1(1/4) = 1 ln⎝1 + 1/4 ⎠ = 1 ln⎝5/4 ⎠ = 1 ln⎝5 ⎠ ≈ 0.2554
2 1 − 1/4
2 3/4
2 3
1.35
Evaluate tanh −1(1/2).
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111
1.5 EXERCISES
For the following exercises, evaluate the given exponential
functions as indicated, accurate to two significant digits
after the decimal.
229.
f (x) = 5 x a. x = 3 b. x = 1 c. x = 2
2
230.
f (x) = (0.3) x a. x = −1 b. x = 4 c. x = −1.5
231.
f (x) = 10 x a. x = −2 b. x = 4 c. x = 5
3
232.
f (x) = e x a. x = 2 b. x = −3.2 c. x = π
234.
For the following exercises, match the exponential equation
to the correct graph.
a.
y = 4 −x
b.
y = 3x − 1
c.
y = 2x + 1
d.
y = ⎝1 ⎠ + 2
2
e.
y = −3 −x
f.
y = 1 − 5x
233.
⎛ ⎞
x
235.
112
236.
Chapter 1 | Functions and Graphs
For the following exercises, sketch the graph of the
exponential function. Determine the domain, range, and
horizontal asymptote.
239.
f (x) = e x + 2
240.
f (x) = −2 x
237.
238.
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113
241.
f (x) = 3 x + 1
243.
f (x) = 1 − 2 −x
242.
f (x) = 4 x − 1
244.
f (x) = 5 x + 1 + 2
114
245.
Chapter 1 | Functions and Graphs
f (x) = e −x − 1
259.
3
64 = 4
260. e x = y
261. 9 y = 150
262. b 3 = 45
263. 4 −3/2 = 0.125
For the following exercises, sketch the graph of the
logarithmic function. Determine the domain, range, and
vertical asymptote.
264.
f (x) = 3 + ln x
265.
f (x) = ln(x − 1)
For the following exercises, write the equation in
equivalent exponential form.
246. log 3 81 = 4
247. log 8 2 = 1
3
248. log 5 1 = 0
249. log 5 25 = 2
250. log0.1 = −1
⎛ ⎞
251. ln 13 = −3
⎝e ⎠
252. log 9 3 = 0.5
253. ln1 = 0
For the following exercises, write the equation in
equivalent logarithmic form.
254. 2 3 = 8
255. 4 −2 = 1
16
256. 10 2 = 100
257. 9 0 = 1
⎛ ⎞
3
258. ⎝1 ⎠ = 1
3
27
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Chapter 1 | Functions and Graphs
115
266.
f (x) = ln(−x)
269.
f (x) = ln(x + 1)
267.
f (x) = 1 − ln x
For the following exercises, use properties of logarithms to
write the expressions as a sum, difference, and/or product
of logarithms.
270. logx 4 y
271. log 3 9a
3
b
3
272. lna b
273. log 5 125xy 3
3
xy
274. log 4
64
⎛
⎞
275. ln⎜ 6 ⎟
268.
f (x) = log x − 1
⎝ e3⎠
For the following exercises, solve the exponential equation
exactly.
276. 5 x = 125
277. e 3x − 15 = 0
278. 8 x = 4
279. 4 x + 1 − 32 = 0
280. 3 x/14 = 1
10
281. 10 x = 7.21
282. 4 · 2 3x − 20 = 0
116
Chapter 1 | Functions and Graphs
283. 7 3x − 2 = 11
For the following exercises, solve the logarithmic equation
exactly, if possible.
284. log 3 x = 0
301. [T] The amount A of a $100,000 investment paying
continuously and compounded for t years is given by
A(t) = 100,000 · e 0.055t. Find the amount A accumulated
in 5 years.
302. [T] An investment is compounded monthly,
quarterly, or yearly and is given by the function
j⎞
⎛
A = P⎝1 + n ⎠ , where A is the value of the investment
285. log 5 x = −2
nt
286. log 4 (x + 5) = 0
at time t, P is the initial principle that was invested, j
is the annual interest rate, and n is the number of time
the interest is compounded per year. Given a yearly interest
rate of 3.5% and an initial principle of $100,000, find the
amount A accumulated in 5 years for interest that is
compounded a. daily, b., monthly, c. quarterly, and d.
yearly.
287. log(2x − 7) = 0
288. ln x + 3 = 2
289. log 6 (x + 9) + log 6 x = 2
290. log 4 (x + 2) − log 4 (x − 1) = 0
303. [T] The concentration of hydrogen ions in a
substance is denoted by ⎡⎣H+ ⎤⎦, measured in moles per
291. ln x + ln(x − 2) = ln4
For the following exercises, use the change-of-base
formula and either base 10 or base e to evaluate the given
expressions. Answer in exact form and in approximate
form, rounding to four decimal places.
292. log 5 47
293. log 7 82
294. log 6 103
liter. The pH of a substance is defined by the logarithmic
function pH = −log⎡⎣H+ ⎤⎦. This function is used to
measure the acidity of a substance. The pH of water is 7. A
substance with a pH less than 7 is an acid, whereas one that
has a pH of more than 7 is a base.
a. Find the pH of the following substances. Round
answers to one digit.
b. Determine whether the substance is an acid or a
base.
i. Eggs: ⎡⎣H+ ⎤⎦ = 1.6 × 10 −8 mol/L
ii. Beer: ⎡⎣H+ ⎤⎦ = 3.16 × 10 −3 mol/L
iii. Tomato Juice: ⎡⎣H+ ⎤⎦ = 7.94 × 10 −5 mol/L
295. log 0.5 211
296. log 2 π
304. [T] Iodine-131 is a radioactive substance that decays
according to the function Q(t) = Q 0 · e −0.08664t, where
297. log 0.2 0.452
Q 0 is the initial quantity of a sample of the substance and t
298. Rewrite the following expressions in terms of
exponentials and simplify. a.
2cosh(ln x)
b.
is in days. Determine how long it takes (to the nearest day)
for 95% of a quantity to decay.
cosh4x + sinh4x
c.
cosh2x − sinh2x
ln(cosh x + sinh x) + ln(cosh x − sinh x)
305. [T] According to the World Bank, at the end of
2013 ( t = 0 ) the U.S. population was 316 million and
was increasing according to the following model:
P(t) = 316e 0.0074t, where P is measured in millions of
d.
299. [T] The number of bacteria N in a culture after t days
can be modeled by the function N(t) = 1300 · (2) t/4. Find
the number of bacteria present after 15 days.
300. [T] The demand D (in millions of barrels) for oil
in an oil-rich country is given by the function
D(p) = 150 · (2.7)
−0.25p
, where p is the price (in dollars)
of a barrel of oil. Find the amount of oil demanded (to the
nearest million barrels) when the price is between $15 and
$20.
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people and t is measured in years after 2013.
a. Based on this model, what will be the population of
the United States in 2020?
b. Determine when the U.S. population will be twice
what it is in 2013.
Chapter 1 | Functions and Graphs
306. [T] The amount A accumulated after 1000 dollars is
invested for t years at an interest rate of 4% is modeled by
the function A(t) = 1000(1.04) t.
a. Find the amount accumulated after 5 years and 10
years.
b. Determine how long it takes for the original
investment to triple.
307. [T] A bacterial colony grown in a lab is known to
double in number in 12 hours. Suppose, initially, there are
1000 bacteria present.
a. Use the exponential function Q = Q 0 e kt to
determine the value k, which is the growth rate of
the bacteria. Round to four decimal places.
b. Determine approximately how long it takes for
200,000 bacteria to grow.
308. [T] The rabbit population on a game reserve doubles
every 6 months. Suppose there were 120 rabbits initially.
a. Use the exponential function P = P 0 a t to
determine the growth rate constant a. Round to
four decimal places.
b. Use the function in part a. to determine
approximately how long it takes for the rabbit
population to reach 3500.
309. [T] The 1906 earthquake in San Francisco had a
magnitude of 8.3 on the Richter scale. At the same time,
in Japan, an earthquake with magnitude 4.9 caused only
minor damage. Approximately how much more energy was
released by the San Francisco earthquake than by the
Japanese earthquake?
117
118
Chapter 1 | Functions and Graphs
CHAPTER 1 REVIEW
KEY TERMS
absolute value function
⎧−x, x < 0
f (x) = ⎨
⎩ x, x ≥ 0
algebraic function a function involving any combination of only the basic operations of addition, subtraction,
multiplication, division, powers, and roots applied to an input variable x
base the number b in the exponential function f (x) = b x and the logarithmic function f (x) = log b x
composite function given two functions f and g, a new function, denoted g ∘ f , such that ⎛⎝g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠
cubic function a polynomial of degree 3; that is, a function of the form f (x) = ax 3 + bx 2 + cx + d, where a ≠ 0
decreasing on the interval I a function decreasing on the interval I if, for all x 1, x 2 ∈ I, f (x 1) ≥ f (x 2) if
x1 < x2
degree for a polynomial function, the value of the largest exponent of any term
dependent variable the output variable for a function
domain the set of inputs for a function
even function a function is even if f (−x) = f (x) for all x in the domain of f
exponent the value x in the expression b x
function a set of inputs, a set of outputs, and a rule for mapping each input to exactly one output
graph of a function the set of points (x, y) such that x is in the domain of f and y = f (x)
horizontal line test a function f is one-to-one if and only if every horizontal line intersects the graph of f , at most,
once
hyperbolic functions the functions denoted sinh, cosh, tanh, csch, sech,
x
combinations of e and e
−x
and coth,
which involve certain
increasing on the interval I a function increasing on the interval I if for all x 1, x 2 ∈ I, f (x 1) ≤ f (x 2) if x 1 < x 2
independent variable the input variable for a function
inverse function for a function f , the inverse function f −1 satisfies f −1 (y) = x if f (x) = y
inverse hyperbolic functions the inverses of the hyperbolic functions where cosh and sech are restricted to the
domain [0, ∞); each of these functions can be expressed in terms of a composition of the natural logarithm function
and an algebraic function
inverse trigonometric functions the inverses of the trigonometric functions are defined on restricted domains where
they are one-to-one functions
linear function a function that can be written in the form f (x) = mx + b
logarithmic function a function of the form f (x) = log b (x) for some base b > 0, b ≠ 1 such that y = log b(x) if
and only if b y = x
mathematical model A method of simulating real-life situations with mathematical equations
natural exponential function the function f (x) = e x
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Chapter 1 | Functions and Graphs
119
natural logarithm the function ln x = log x
e
number e as m gets larger, the quantity (1 + (1/m) m gets closer to some real number; we define that real number to be
e; the value of e is approximately 2.718282
odd function a function is odd if f (−x) = − f (x) for all x in the domain of f
one-to-one function a function f is one-to-one if f (x 1) ≠ f (x 2) if x 1 ≠ x 2
periodic function a function is periodic if it has a repeating pattern as the values of x move from left to right
piecewise-defined function a function that is defined differently on different parts of its domain
point-slope equation equation of a linear function indicating its slope and a point on the graph of the function
n−1
polynomial function a function of the form f (x) = a n x n + a
+ … + a1 x + a0
n−1x
power function a function of the form f (x) = x n for any positive integer n ≥ 1
quadratic function a polynomial of degree 2; that is, a function of the form f (x) = ax 2 + bx + c where a ≠ 0
radians for a circular arc of length s on a circle of radius 1, the radian measure of the associated angle θ is s
range the set of outputs for a function
rational function a function of the form f (x) = p(x)/q(x), where p(x) and q(x) are polynomials
restricted domain a subset of the domain of a function f
root function a function of the form f (x) = x 1/n for any integer n ≥ 2
slope the change in y for each unit change in x
slope-intercept form equation of a linear function indicating its slope and y-intercept
symmetry about the origin the graph of a function f is symmetric about the origin if (−x, −y) is on the graph of f
whenever (x, y) is on the graph
symmetry about the y-axis the graph of a function f is symmetric about the y -axis if (−x, y) is on the graph of f
whenever (x, y) is on the graph
table of values a table containing a list of inputs and their corresponding outputs
transcendental function a function that cannot be expressed by a combination of basic arithmetic operations
transformation of a function a shift, scaling, or reflection of a function
trigonometric functions functions of an angle defined as ratios of the lengths of the sides of a right triangle
trigonometric identity an equation involving trigonometric functions that is true for all angles θ for which the
functions in the equation are defined
vertical line test given the graph of a function, every vertical line intersects the graph, at most, once
zeros of a function when a real number x is a zero of a function f , f (x) = 0
KEY EQUATIONS
• Composition of two functions
⎛
⎝
g ∘ f ⎞⎠(x) = g⎛⎝ f (x)⎞⎠
• Absolute value function
⎧−x, x < 0
f (x) = ⎨
⎩ x, x ≥ 0
120
Chapter 1 | Functions and Graphs
• Point-slope equation of a line
y − y 1 = m(x − x 1)
• Slope-intercept form of a line
y = mx + b
• Standard form of a line
ax + by = c
• Polynomial function
f (x) = a n x n + a n − 1 x n − 1 + ⋯ + a 1 x + a 0
• Generalized sine function
f (x) = Asin⎛⎝B(x − α)⎞⎠ + C
• Inverse functions
f −1 ⎛⎝ f (x)⎞⎠ = x for all x in D, and f ⎛⎝ f −1 (y)⎞⎠ = y for all y in R.
KEY CONCEPTS
1.1 Review of Functions
• A function is a mapping from a set of inputs to a set of outputs with exactly one output for each input.
• If no domain is stated for a function y = f (x), the domain is considered to be the set of all real numbers x for
which the function is defined.
• When sketching the graph of a function f , each vertical line may intersect the graph, at most, once.
• A function may have any number of zeros, but it has, at most, one y-intercept.
• To define the composition g ∘ f , the range of f must be contained in the domain of g.
• Even functions are symmetric about the y -axis whereas odd functions are symmetric about the origin.
1.2 Basic Classes of Functions
• The power function f (x) = x n is an even function if n is even and n ≠ 0, and it is an odd function if n is odd.
• The root function f (x) = x 1/n has the domain [0, ∞) if n is even and the domain (−∞, ∞) if n is odd. If n
is odd, then f (x) = x 1/n is an odd function.
• The domain of the rational function f (x) = p(x)/q(x), where p(x) and q(x) are polynomial functions, is the set
of x such that q(x) ≠ 0.
• Functions that involve the basic operations of addition, subtraction, multiplication, division, and powers are
algebraic functions. All other functions are transcendental. Trigonometric, exponential, and logarithmic functions
are examples of transcendental functions.
• A polynomial function f with degree n ≥ 1 satisfies f (x) → ±∞ as x → ±∞. The sign of the output as
x → ∞ depends on the sign of the leading coefficient only and on whether n is even or odd.
• Vertical and horizontal shifts, vertical and horizontal scalings, and reflections about the x - and y -axes are
examples of transformations of functions.
1.3 Trigonometric Functions
• Radian measure is defined such that the angle associated with the arc of length 1 on the unit circle has radian
measure 1. An angle with a degree measure of 180° has a radian measure of π rad.
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Chapter 1 | Functions and Graphs
121
• For acute angles θ, the values of the trigonometric functions are defined as ratios of two sides of a right triangle
in which one of the acute angles is θ.
• For a general angle θ,
let (x, y) be a point on a circle of radius r corresponding to this angle θ. The
trigonometric functions can be written as ratios involving x, y, and r.
• The trigonometric functions are periodic. The sine, cosine, secant, and cosecant functions have period 2π. The
tangent and cotangent functions have period π.
1.4 Inverse Functions
• For a function to have an inverse, the function must be one-to-one. Given the graph of a function, we can determine
whether the function is one-to-one by using the horizontal line test.
• If a function is not one-to-one, we can restrict the domain to a smaller domain where the function is one-to-one and
then define the inverse of the function on the smaller domain.
• For a function f and its inverse f −1, f ⎛⎝ f −1 (x)⎞⎠ = x for all x in the domain of f −1 and f −1 ⎛⎝ f (x)⎞⎠ = x for all
x in the domain of f .
• Since the trigonometric functions are periodic, we need to restrict their domains to define the inverse trigonometric
functions.
• The graph of a function f and its inverse f −1 are symmetric about the line y = x.
1.5 Exponential and Logarithmic Functions
• The exponential function y = b x is increasing if b > 1 and decreasing if 0 < b < 1. Its domain is (−∞, ∞)
and its range is (0, ∞).
• The logarithmic function y = log b(x) is the inverse of y = b x. Its domain is (0, ∞) and its range is (−∞, ∞).
• The natural exponential function is y = e x and the natural logarithmic function is y = ln x = log e x.
• Given an exponential function or logarithmic function in base a, we can make a change of base to convert this
function to any base b > 0, b ≠ 1. We typically convert to base e.
• The hyperbolic functions involve combinations of the exponential functions e x and e −x. As a result, the inverse
hyperbolic functions involve the natural logarithm.
CHAPTER 1 REVIEW EXERCISES
True or False? Justify your answer with a proof or a
counterexample.
310. A function is always one-to-one.
311.
f ∘ g = g ∘ f , assuming f and g are functions.
312. A relation that passes the horizontal and vertical line
tests is a one-to-one function.
313. A relation passing the horizontal line test is a
function.
For the following problems, state the domain and range of
the given functions:
f = x 2 + 2x − 3,
g = ln(x − 5),
h=
1
x+4
314. h
315. g
316. h ∘ f
317. g ∘ f
Find the degree, y-intercept, and zeros for the following
polynomial functions.
318.
f (x) = 2x 2 + 9x − 5
122
319.
Chapter 1 | Functions and Graphs
f (x) = x 3 + 2x 2 − 2x
Simplify the following trigonometric expressions.
320.
tan 2 x + cos 2 x
sec 2 x
321. cos(2x) = sin 2 x
331. a. Find the equation C = f (x) that describes the
total cost as a function of number of shirts and b. determine
how many shirts he must sell to break even if he sells the
shirts for $10 each.
332.
a. Find the inverse function x = f −1(C) and
describe the meaning of this function. b. Determine how
many shirts the owner can buy if he has $8000 to spend.
Solve the following trigonometric equations on the interval
θ = [−2π, 2π] exactly.
For the following problems, consider the population of
Ocean City, New Jersey, which is cyclical by season.
322. 6cos 2 x − 3 = 0
P(t) = 82.5 − 67.5cos⎡⎣(π/6)t⎤⎦,
2
323. sec x − 2sec x + 1 = 0
Solve the following logarithmic equations.
324. 5 x = 16
325. log 2(x + 4) = 3
Are the following functions one-to-one over their domain
of existence? Does the function have an inverse? If so, find
the inverse f −1(x) of the function. Justify your answer.
326.
f (x) = x 2 + 2x + 1
327.
f (x) = 1x
For the following problems, determine the largest domain
on which the function is one-to-one and find the inverse on
that domain.
328.
f (x) = 9 − x
329.
f (x) = x 2 + 3x + 4
330. A car is racing along a circular track with diameter of
1 mi. A trainer standing in the center of the circle marks his
progress every 5 sec. After 5 sec, the trainer has to turn 55°
to keep up with the car. How fast is the car traveling?
For the following problems, consider a restaurant owner
who wants to sell T-shirts advertising his brand. He recalls
that there is a fixed cost and variable cost, although he does
not remember the values. He does know that the T-shirt
printing company charges $440 for 20 shirts and $1000 for
100 shirts.
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333.
The
population
can
be
where
t
modeled by
is time in
months (t = 0 represents January 1) and P is population
(in thousands). During a year, in what intervals is the
population less than 20,000? During what intervals is the
population more than 140,000?
334. In reality, the overall population is most likely
increasing or decreasing throughout each year. Let’s
reformulate
the
model
as
P(t) = 82.5 − 67.5cos⎡⎣(π/6)t⎤⎦ + t, where t is time in
months ( t = 0 represents January 1) and P is population
(in thousands). When is the first time the population
reaches 200,000?
For the following problems, consider radioactive dating. A
human skeleton is found in an archeological dig. Carbon
dating is implemented to determine how old the skeleton is
by using the equation y = e rt, where y is the percentage
of radiocarbon still present in the material, t is the number
of years passed, and r = −0.0001210 is the decay rate of
radiocarbon.
335. If the skeleton is expected to be 2000 years old, what
percentage of radiocarbon should be present?
336. Find the inverse of the carbon-dating equation. What
does it mean? If there is 25% radiocarbon, how old is the
skeleton?
Chapter 2 | Limits
123
2 | LIMITS
Figure 2.1 The vision of human exploration by the National Aeronautics and Space Administration (NASA) to distant parts of
the universe illustrates the idea of space travel at high speeds. But, is there a limit to how fast a spacecraft can go? (credit:
NASA)
Chapter Outline
2.1 A Preview of Calculus
2.2 The Limit of a Function
2.3 The Limit Laws
2.4 Continuity
2.5 The Precise Definition of a Limit
Introduction
Science fiction writers often imagine spaceships that can travel to far-off planets in distant galaxies. However, back in 1905,
Albert Einstein showed that a limit exists to how fast any object can travel. The problem is that the faster an object moves,
the more mass it attains (in the form of energy), according to the equation
m=
m0
2
1 − v2
c
,
124
Chapter 2 | Limits
where m0 is the object’s mass at rest, v is its speed, and c is the speed of light. What is this speed limit? (We explore this
problem further in Example 2.12.)
The idea of a limit is central to all of calculus. We begin this chapter by examining why limits are so important. Then, we
go on to describe how to find the limit of a function at a given point. Not all functions have limits at all points, and we
discuss what this means and how we can tell if a function does or does not have a limit at a particular value. This chapter has
been created in an informal, intuitive fashion, but this is not always enough if we need to prove a mathematical statement
involving limits. The last section of this chapter presents the more precise definition of a limit and shows how to prove
whether a function has a limit.
2.1 | A Preview of Calculus
Learning Objectives
2.1.1 Describe the tangent problem and how it led to the idea of a derivative.
2.1.2 Explain how the idea of a limit is involved in solving the tangent problem.
2.1.3 Recognize a tangent to a curve at a point as the limit of secant lines.
2.1.4 Identify instantaneous velocity as the limit of average velocity over a small time interval.
2.1.5 Describe the area problem and how it was solved by the integral.
2.1.6 Explain how the idea of a limit is involved in solving the area problem.
2.1.7 Recognize how the ideas of limit, derivative, and integral led to the studies of infinite series
and multivariable calculus.
As we embark on our study of calculus, we shall see how its development arose from common solutions to practical
problems in areas such as engineering physics—like the space travel problem posed in the chapter opener. Two key
problems led to the initial formulation of calculus: (1) the tangent problem, or how to determine the slope of a line tangent
to a curve at a point; and (2) the area problem, or how to determine the area under a curve.
The Tangent Problem and Differential Calculus
Rate of change is one of the most critical concepts in calculus. We begin our investigation of rates of change by looking at
the graphs of the three lines f (x) = −2x − 3, g(x) = 1 x + 1, and h(x) = 2, shown in Figure 2.2.
2
Figure 2.2 The rate of change of a linear function is constant in each of these three graphs, with the constant determined by the
slope.
As we move from left to right along the graph of f (x) = −2x − 3, we see that the graph decreases at a constant rate. For
every 1 unit we move to the right along the x-axis, the y-coordinate decreases by 2 units. This rate of change is determined
by the slope (−2) of the line. Similarly, the slope of 1/2 in the function g(x) tells us that for every change in x of 1 unit
there is a corresponding change in y of 1/2 unit. The function h(x) = 2 has a slope of zero, indicating that the values of the
function remain constant. We see that the slope of each linear function indicates the rate of change of the function.
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Chapter 2 | Limits
125
Compare the graphs of these three functions with the graph of k(x) = x 2 (Figure 2.3). The graph of k(x) = x 2 starts from
the left by decreasing rapidly, then begins to decrease more slowly and level off, and then finally begins to increase—slowly
at first, followed by an increasing rate of increase as it moves toward the right. Unlike a linear function, no single number
represents the rate of change for this function. We quite naturally ask: How do we measure the rate of change of a nonlinear
function?
Figure 2.3 The function k(x) = x 2 does not have a constant
rate of change.
We can approximate the rate of change of a function f (x) at a point ⎛⎝a, f (a)⎞⎠ on its graph by taking another point ⎛⎝x, f (x)⎞⎠
on the graph of f (x), drawing a line through the two points, and calculating the slope of the resulting line. Such a line is
called a secant line. Figure 2.4 shows a secant line to a function f (x) at a point ⎛⎝a, f (a)⎞⎠.
Figure 2.4 The slope of a secant line through a point
⎛
⎞
⎝a, f (a)⎠ estimates the rate of change of the function at the
point ⎛⎝a, f (a)⎞⎠.
We formally define a secant line as follows:
126
Chapter 2 | Limits
Definition
The secant to the function f (x) through the points ⎛⎝a, f (a)⎞⎠ and ⎛⎝x, f (x)⎞⎠ is the line passing through these points. Its
slope is given by
m sec =
f (x) − f (a)
x−a .
(2.1)
The accuracy of approximating the rate of change of the function with a secant line depends on how close x is to a. As we
see in Figure 2.5, if x is closer to a, the slope of the secant line is a better measure of the rate of change of f (x) at a.
Figure 2.5 As x gets closer to a, the slope of the secant line
becomes a better approximation to the rate of change of the
function f (x) at a.
The secant lines themselves approach a line that is called the tangent to the function f (x) at a (Figure 2.6). The slope of
the tangent line to the graph at a measures the rate of change of the function at a. This value also represents the derivative of
the function f (x) at a, or the rate of change of the function at a. This derivative is denoted by f ′ (a). Differential calculus
is the field of calculus concerned with the study of derivatives and their applications.
For an interactive demonstration of the slope of a secant line that you can manipulate yourself, visit this
applet (Note: this site requires a Java browser plugin): Math Insight (http://www.openstaxcollege.org/l/
20_mathinsight) .
Figure 2.6 Solving the Tangent Problem: As x approaches a,
the secant lines approach the tangent line.
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Chapter 2 | Limits
127
Example 2.1 illustrates how to find slopes of secant lines. These slopes estimate the slope of the tangent line or,
equivalently, the rate of change of the function at the point at which the slopes are calculated.
Example 2.1
Finding Slopes of Secant Lines
Estimate the slope of the tangent line (rate of change) to f (x) = x 2 at x = 1 by finding slopes of secant lines
through (1, 1) and each of the following points on the graph of f (x) = x 2.
a.
(2, 4)
b.
⎛3 9 ⎞
⎝2 , 4 ⎠
Solution
Use the formula for the slope of a secant line from the definition.
a.
m sec = 4 − 1 = 3
2−1
b.
m sec =
9
4
3
2
−1
−1
= 5 = 2.5
2
The point in part b. is closer to the point (1, 1), so the slope of 2.5 is closer to the slope of the tangent line. A
good estimate for the slope of the tangent would be in the range of 2 to 2.5 (Figure 2.7).
Figure 2.7 The secant lines to f (x) = x 2 at (1, 1) through
⎛
⎞
(a) (2, 4) and (b) ⎝3 , 9 ⎠ provide successively closer
2 4
approximations to the tangent line to f (x) = x 2 at (1, 1).
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Chapter 2 | Limits
2.1
Estimate the slope of the tangent line (rate of change) to f (x) = x 2 at x = 1 by finding slopes of secant
⎛
⎞
lines through (1, 1) and the point ⎝5 , 25 ⎠ on the graph of f (x) = x 2.
4 16
We continue our investigation by exploring a related question. Keeping in mind that velocity may be thought of as the rate
of change of position, suppose that we have a function, s(t), that gives the position of an object along a coordinate axis
at any given time t. Can we use these same ideas to create a reasonable definition of the instantaneous velocity at a given
time t = a ? We start by approximating the instantaneous velocity with an average velocity. First, recall that the speed of
an object traveling at a constant rate is the ratio of the distance traveled to the length of time it has traveled. We define the
average velocity of an object over a time period to be the change in its position divided by the length of the time period.
Definition
Let s(t) be the position of an object moving along a coordinate axis at time t. The average velocity of the object over
a time interval [a, t] where a < t (or [t, a] if t < a) is
v ave =
s(t) − s(a)
t−a .
(2.2)
As t is chosen closer to a, the average velocity becomes closer to the instantaneous velocity. Note that finding the average
velocity of a position function over a time interval is essentially the same as finding the slope of a secant line to a function.
Furthermore, to find the slope of a tangent line at a point a, we let the x-values approach a in the slope of the secant line.
Similarly, to find the instantaneous velocity at time a, we let the t-values approach a in the average velocity. This process
of letting x or t approach a in an expression is called taking a limit. Thus, we may define the instantaneous velocity as
follows.
Definition
For a position function s(t), the instantaneous velocity at a time t = a is the value that the average velocities
approach on intervals of the form [a, t] and [t, a] as the values of t become closer to a, provided such a value exists.
Example 2.2 illustrates this concept of limits and average velocity.
Example 2.2
Finding Average Velocity
A rock is dropped from a height of 64 ft. It is determined that its height (in feet) above ground t seconds later (for
0 ≤ t ≤ 2) is given by s(t) = −16t 2 + 64. Find the average velocity of the rock over each of the given time
intervals. Use this information to guess the instantaneous velocity of the rock at time t = 0.5.
a.
⎡
⎣
0.49, 0.5⎤⎦
b.
⎡
⎣
0.5, 0.51⎤⎦
Solution
Substitute the data into the formula for the definition of average velocity.
a.
v ave =
s(0.49) − s(0.5)
= −15.84
0.49 − 0.5
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Chapter 2 | Limits
b.
v ave =
129
s(0.51) − s(0.5)
= −16.016
0.51 − 0.5
The instantaneous velocity is somewhere between −15.84 and −16.16 ft/sec. A good guess might be −16 ft/sec.
2.2
An object moves along a coordinate axis so that its position at time t is given by s(t) = t 3. Estimate its
instantaneous velocity at time t = 2 by computing its average velocity over the time interval [2, 2.001].
The Area Problem and Integral Calculus
We now turn our attention to a classic question from calculus. Many quantities in physics—for example, quantities of
work—may be interpreted as the area under a curve. This leads us to ask the question: How can we find the area between
the graph of a function and the x-axis over an interval (Figure 2.8)?
Figure 2.8 The Area Problem: How do we find the area of the
shaded region?
As in the answer to our previous questions on velocity, we first try to approximate the solution. We approximate the area by
dividing up the interval ⎡⎣a, b⎤⎦ into smaller intervals in the shape of rectangles. The approximation of the area comes from
adding up the areas of these rectangles (Figure 2.9).
Figure 2.9 The area of the region under the curve is
approximated by summing the areas of thin rectangles.
As the widths of the rectangles become smaller (approach zero), the sums of the areas of the rectangles approach the area
between the graph of f (x) and the x-axis over the interval ⎡⎣a, b⎤⎦. Once again, we find ourselves taking a limit. Limits
of this type serve as a basis for the definition of the definite integral. Integral calculus is the study of integrals and their
applications.
130
Chapter 2 | Limits
Example 2.3
Estimation Using Rectangles
Estimate the area between the x-axis and the graph of f (x) = x 2 + 1 over the interval [0, 3] by using the three
rectangles shown in Figure 2.10.
Figure 2.10 The area of the region under the curve of
f (x) = x 2 + 1 can be estimated using rectangles.
Solution
The areas of the three rectangles are 1 unit2, 2 unit2, and 5 unit2. Using these rectangles, our area estimate is 8
unit2.
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Chapter 2 | Limits
2.3
131
Estimate the area between the x-axis and the graph of f (x) = x 2 + 1 over the interval [0, 3] by using
the three rectangles shown here:
Other Aspects of Calculus
So far, we have studied functions of one variable only. Such functions can be represented visually using graphs in two
dimensions; however, there is no good reason to restrict our investigation to two dimensions. Suppose, for example, that
instead of determining the velocity of an object moving along a coordinate axis, we want to determine the velocity of a
rock fired from a catapult at a given time, or of an airplane moving in three dimensions. We might want to graph real-value
functions of two variables or determine volumes of solids of the type shown in Figure 2.11. These are only a few of the
types of questions that can be asked and answered using multivariable calculus. Informally, multivariable calculus can be
characterized as the study of the calculus of functions of two or more variables. However, before exploring these and other
ideas, we must first lay a foundation for the study of calculus in one variable by exploring the concept of a limit.
Figure 2.11 We can use multivariable calculus to find the
volume between a surface defined by a function of two variables
and a plane.
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Chapter 2 | Limits
2.1 EXERCISES
For the following exercises, points P(1, 2) and Q(x, y)
are on the graph of the function f (x) = x 2 + 1.
5. Use the values in the right column of the table in the
preceding exercise to guess the value of the slope of the
tangent line to f at x = 1.
1. [T] Complete the following table with the appropriate
values: y-coordinate of Q, the point Q(x, y), and the slope
6. Use the value in the preceding exercise to find the
equation of the tangent line at point P. Graph f (x) and the
of the secant line passing through points P and Q. Round
your answer to eight significant digits.
tangent line.
For the following exercises, points P(4, 2) and Q(x, y)
x
y
Q(x, y)
msec
1.1
a.
e.
i.
1.01
b.
f.
j.
1.001
c.
g.
k.
x
y
Q(x, y)
msec
1.0001
d.
h.
l.
4.1
a.
e.
i.
4.01
b.
f.
j.
4.001
c.
g.
k.
4.0001
d.
h.
l.
2. Use the values in the right column of the table in the
preceding exercise to guess the value of the slope of the line
tangent to f at x = 1.
3. Use the value in the preceding exercise to find the
equation of the tangent line at point P. Graph f (x) and the
are on the graph of the function f (x) = x.
7. [T] Complete the following table with the appropriate
values: y-coordinate of Q, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Round
your answer to eight significant digits.
tangent line.
are on the graph of the function f (x) = x 3.
8. Use the values in the right column of the table in the
preceding exercise to guess the value of the slope of the
tangent line to f at x = 4.
4. [T] Complete the following table with the appropriate
values: y-coordinate of Q, the point Q(x, y), and the slope
9. Use the value in the preceding exercise to find the
equation of the tangent line at point P.
of the secant line passing through points P and Q. Round
your answer to eight significant digits.
For the following exercises, points P(1.5, 0) and Q⎛⎝ϕ, y⎞⎠
For the following exercises, points P(1, 1) and Q(x, y)
x
y
Q(x, y)
msec
1.1
a.
e.
i.
1.01
b.
f.
j.
1.001
c.
g.
k.
1.0001
d.
h.
l.
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are on the graph of the function f ⎛⎝ϕ⎞⎠ = cos ⎛⎝πϕ⎞⎠.
Chapter 2 | Limits
133
10. [T] Complete the following table with the appropriate
values: y-coordinate of Q, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Round
your answer to eight significant digits.
x
y
Q⎛⎝ϕ, y⎞⎠
msec
1.4
a.
e.
i.
by s(t) = 200 − 4.9t 2, where position s is measured in
meters and time t is measured in seconds. Round your
answer to eight significant digits.
16. [T] Compute the average velocity of the ball over the
given time intervals.
a. ⎡⎣4.99, 5⎤⎦
b.
⎡
⎣
5, 5.01⎤⎦
c.
⎡
⎣
d.
⎡
⎣
4.999, 5⎤⎦
1.49
b.
f.
j.
1.499
c.
g.
k.
17. Use the preceding exercise to guess the instantaneous
velocity of the ball at t = 5 sec.
1.4999
d.
h.
l.
For the following exercises, consider a stone tossed into the
air from ground level with an initial velocity of 15 m/sec.
Its height in meters at time t seconds is h(t) = 15t − 4.9t 2.
11. Use the values in the right column of the table in the
preceding exercise to guess the value of the slope of the
tangent line to f at x = 4.
5, 5.001⎤⎦
18. [T] Compute the average velocity of the stone over the
given time intervals.
a. ⎡⎣1, 1.05⎤⎦
12. Use the value in the preceding exercise to find the
equation of the tangent line at point P.
b.
c.
[1, 1.01]
⎡
⎤
⎣1, 1.005⎦
For the following exercises, points P(−1, −1) and
d.
[1, 1.001]
Q(x, y) are on the graph of the function f (x) = 1x .
13. [T] Complete the following table with the appropriate
values: y-coordinate of Q, the point Q(x, y), and the slope
of the secant line passing through points P and Q. Round
your answer to eight significant digits.
x
y
Q(x, y)
msec
−1.05
a.
e.
i.
−1.01
b.
f.
j.
−1.005
c.
g.
k.
−1.001
d.
h.
l.
19. Use the preceding exercise to guess the instantaneous
velocity of the stone at t = 1 sec.
For the following exercises, consider a rocket shot into the
air that then returns to Earth. The height of the rocket in
meters is given by h(t) = 600 + 78.4t − 4.9t 2, where t is
measured in seconds.
20. [T] Compute the average velocity of the rocket over
the given time intervals.
a. [9, 9.01]
b.
c.
d.
[8.99, 9]
[9, 9.001]
[8.999, 9]
21. Use the preceding exercise to guess the instantaneous
velocity of the rocket at t = 9 sec.
For the following exercises, consider an athlete running
a 40-m dash. The position of the athlete is given by
14. Use the values in the right column of the table in the
preceding exercise to guess the value of the slope of the line
tangent to f at x = −1.
15. Use the value in the preceding exercise to find the
equation of the tangent line at point P.
For the following exercises, the position function of a ball
dropped from the top of a 200-meter tall building is given
3
d(t) = t + 4t, where d is the position in meters and t is
6
the time elapsed, measured in seconds.
134
Chapter 2 | Limits
22. [T] Compute the average velocity of the runner over
the given time intervals.
a. ⎡⎣1.95, 2.05⎤⎦
b.
⎡
⎣
c.
⎡
⎣
1.995, 2.005⎤⎦
d.
[2, 2.00001]
1.9995, 2.0005⎤⎦
23. Use the preceding exercise to guess the instantaneous
velocity of the runner at t = 2 sec.
For the following exercises, consider the function
f (x) = |x|.
24. Sketch the graph of f over the interval [−1, 2] and
shade the region above the x-axis.
25. Use the preceding exercise to find the exact value of
the area between the x-axis and the graph of f over the
interval [−1, 2] using rectangles. For the rectangles, use
the square units, and approximate both above and below the
lines. Use geometry to find the exact answer.
For the following exercises, consider the function
f (x) = 1 − x 2. (Hint: This is the upper half of a circle of
radius 1 positioned at (0, 0).)
26. Sketch the graph of f over the interval [−1, 1].
27. Use the preceding exercise to find the exact area
between the x-axis and the graph of f over the interval
[−1, 1] using rectangles. For the rectangles, use squares
0.4 by 0.4 units, and approximate both above and below the
lines. Use geometry to find the exact answer.
For the following exercises, consider the function
f (x) = −x 2 + 1.
28. Sketch the graph of f over the interval [−1, 1].
29. Approximate the area of the region between the x-axis
and the graph of f over the interval [−1, 1].
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Chapter 2 | Limits
135
2.2 | The Limit of a Function
Learning Objectives
2.2.1
2.2.2
exist.
2.2.3
2.2.4
2.2.5
2.2.6
2.2.7
Using correct notation, describe the limit of a function.
Use a table of values to estimate the limit of a function or to identify when the limit does not
Use a graph to estimate the limit of a function or to identify when the limit does not exist.
Define one-sided limits and provide examples.
Explain the relationship between one-sided and two-sided limits.
Using correct notation, describe an infinite limit.
Define a vertical asymptote.
The concept of a limit or limiting process, essential to the understanding of calculus, has been around for thousands of years.
In fact, early mathematicians used a limiting process to obtain better and better approximations of areas of circles. Yet, the
formal definition of a limit—as we know and understand it today—did not appear until the late 19th century. We therefore
begin our quest to understand limits, as our mathematical ancestors did, by using an intuitive approach. At the end of this
chapter, armed with a conceptual understanding of limits, we examine the formal definition of a limit.
We begin our exploration of limits by taking a look at the graphs of the functions
2
1 ,
f (x) = x − 4 , g(x) = |x − 2| , and h(x) =
x−2
x−2
(x − 2) 2
which are shown in Figure 2.12. In particular, let’s focus our attention on the behavior of each graph at and around x = 2.
Figure 2.12 These graphs show the behavior of three different functions around x = 2.
Each of the three functions is undefined at x = 2, but if we make this statement and no other, we give a very incomplete
picture of how each function behaves in the vicinity of x = 2. To express the behavior of each graph in the vicinity of 2
more completely, we need to introduce the concept of a limit.
Intuitive Definition of a Limit
Let’s first take a closer look at how the function f (x) = (x 2 − 4)/(x − 2) behaves around x = 2 in Figure 2.12. As the
values of x approach 2 from either side of 2, the values of y = f (x) approach 4. Mathematically, we say that the limit of
f (x) as x approaches 2 is 4. Symbolically, we express this limit as
136
Chapter 2 | Limits
lim f (x) = 4.
x→2
From this very brief informal look at one limit, let’s start to develop an intuitive definition of the limit. We can think of the
limit of a function at a number a as being the one real number L that the functional values approach as the x-values approach
a, provided such a real number L exists. Stated more carefully, we have the following definition:
Definition
Let f (x) be a function defined at all values in an open interval containing a, with the possible exception of a itself,
and let L be a real number. If all values of the function f (x) approach the real number L as the values of x( ≠ a)
approach the number a, then we say that the limit of f (x) as x approaches a is L. (More succinct, as x gets closer to a,
f (x) gets closer and stays close to L.) Symbolically, we express this idea as
lim f (x) = L.
(2.3)
x→a
We can estimate limits by constructing tables of functional values and by looking at their graphs. This process is described
in the following Problem-Solving Strategy.
Problem-Solving Strategy: Evaluating a Limit Using a Table of Functional Values
1. To evaluate xlim
f (x), we begin by completing a table of functional values. We should choose two sets of
→a
x-values—one set of values approaching a and less than a, and another set of values approaching a and greater
than a. Table 2.1 demonstrates what your tables might look like.
x
f(x)
x
f(x)
a − 0.1
f (a − 0.1)
a + 0.1
f (a + 0.1)
a − 0.01
f (a − 0.01)
a + 0.01
f (a + 0.01)
a − 0.001
f (a − 0.001)
a + 0.001
f (a + 0.001)
a − 0.0001
f (a − 0.0001)
a + 0.0001
f (a + 0.0001)
Use additional values as necessary.
Use additional values as necessary.
f (x)
Table 2.1 Table of Functional Values for xlim
→a
2. Next, let’s look at the values in each of the f (x) columns and determine whether the values seem to
be approaching a single value as we move down each column. In our columns, we look at the sequence
f (a − 0.1), f (a − 0.01), f (a − 0.001)., f (a − 0.0001),
and
so
on,
and
f (a + 0.1), f (a + 0.01), f (a + 0.001), f (a + 0.0001), and so on. (Note: Although we have chosen the
x-values a ± 0.1, a ± 0.01, a ± 0.001, a ± 0.0001, and so forth, and these values will probably work nearly
every time, on very rare occasions we may need to modify our choices.)
f (x) = L. We can use the following strategy to
3. If both columns approach a common y-value L, we state xlim
→a
confirm the result obtained from the table or as an alternative method for estimating a limit.
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Chapter 2 | Limits
137
4. Using a graphing calculator or computer software that allows us graph functions, we can plot the function
f (x), making sure the functional values of f (x) for x-values near a are in our window. We can use the trace
feature to move along the graph of the function and watch the y-value readout as the x-values approach a. If
the y-values approach L as our x-values approach a from both directions, then xlim
f (x) = L. We may need
→a
to zoom in on our graph and repeat this process several times.
We apply this Problem-Solving Strategy to compute a limit in Example 2.4.
Example 2.4
Evaluating a Limit Using a Table of Functional Values 1
x
Evaluate lim sin
x using a table of functional values.
x→0
Solution
We have calculated the values of f (x) = (sin x)/x for the values of x listed in Table 2.2.
x
sinx
x
x
sinx
x
−0.1
0.998334166468
0.1
0.998334166468
−0.01
0.999983333417
0.01
0.999983333417
−0.001
0.999999833333
0.001
0.999999833333
−0.0001
0.999999998333
0.0001
0.999999998333
Table 2.2
x
Table of Functional Values for lim sin
x
x→0
Note: The values in this table were obtained using a calculator and using all the places given in the calculator
output.
(sin x)
column, we see that the values in each column appear to be approaching
x
x = 1. A calculator-or computer-generated graph of
one. Thus, it is fairly reasonable to conclude that lim sin
x→0 x
As we read down each
f (x) =
(sin x)
x would be similar to that shown in Figure 2.13, and it confirms our estimate.
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Chapter 2 | Limits
Figure 2.13 The graph of f (x) = (sin x)/x confirms the
estimate from Table 2.2.
Example 2.5
Evaluating a Limit Using a Table of Functional Values 2
Evaluate lim x − 2 using a table of functional values.
x→4
x−4
Solution
As before, we use a table—in this case, Table 2.3—to list the values of the function for the given values of x.
x
x−2
x−4
x
x−2
x−4
3.9
0.251582341869
4.1
0.248456731317
3.99
0.25015644562
4.01
0.24984394501
3.999
0.250015627
4.001
0.249984377
3.9999
0.250001563
4.0001
0.249998438
3.99999
0.25000016
4.00001
0.24999984
Table 2.3
Table of Functional Values for lim x − 2
x→4
x−4
After inspecting this table, we see that the functional values less than 4 appear to be decreasing toward
0.25 whereas the functional values greater than 4 appear to be increasing toward 0.25. We conclude that
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Chapter 2 | Limits
139
lim x − 2 = 0.25. We confirm this estimate using the graph of f (x) = x − 2 shown in Figure 2.14.
x−4
x−4
x→4
Figure 2.14 The graph of f (x) = x − 2 confirms the
x−4
estimate from Table 2.3.
2.4
1
−1
Estimate lim x
using a table of functional values. Use a graph to confirm your estimate.
x → 1x − 1
At this point, we see from Example 2.4 and Example 2.5 that it may be just as easy, if not easier, to estimate a limit of
a function by inspecting its graph as it is to estimate the limit by using a table of functional values. In Example 2.6, we
evaluate a limit exclusively by looking at a graph rather than by using a table of functional values.
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Chapter 2 | Limits
Example 2.6
Evaluating a Limit Using a Graph
For g(x) shown in Figure 2.15, evaluate
lim g(x).
x → −1
Figure 2.15 The graph of g(x) includes one value not on a
smooth curve.
Solution
Despite the fact that g(−1) = 4, as the x-values approach −1 from either side, the g(x) values approach 3.
Therefore,
lim g(x) = 3. Note that we can determine this limit without even knowing the algebraic expression
x → −1
of the function.
Based on Example 2.6, we make the following observation: It is possible for the limit of a function to exist at a point, and
for the function to be defined at this point, but the limit of the function and the value of the function at the point may be
different.
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Chapter 2 | Limits
2.5
141
Use the graph of h(x) in Figure 2.16 to evaluate lim h(x), if possible.
x→2
Figure 2.16
Looking at a table of functional values or looking at the graph of a function provides us with useful insight into the value
of the limit of a function at a given point. However, these techniques rely too much on guesswork. We eventually need to
develop alternative methods of evaluating limits. These new methods are more algebraic in nature and we explore them in
the next section; however, at this point we introduce two special limits that are foundational to the techniques to come.
Theorem 2.1: Two Important Limits
Let a be a real number and c be a constant.
i.
ii.
lim x = a
(2.4)
lim c = c
(2.5)
x→a
x→a
We can make the following observations about these two limits.
i. For the first limit, observe that as x approaches a, so does f (x), because f (x) = x. Consequently, xlim
x = a.
→a
ii. For the second limit, consider Table 2.4.
x
f(x) = c
x
f(x) = c
a − 0.1
c
a + 0.1
c
a − 0.01
c
a + 0.01
c
a − 0.001
c
a + 0.001
c
a − 0.0001
c
a + 0.0001
c
Table 2.4 Table of Functional Values for xlim
c=c
→a
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Chapter 2 | Limits
Observe that for all values of x (regardless of whether they are approaching a), the values f (x) remain constant at c. We
have no choice but to conclude xlim
c = c.
→a
The Existence of a Limit
As we consider the limit in the next example, keep in mind that for the limit of a function to exist at a point, the functional
values must approach a single real-number value at that point. If the functional values do not approach a single value, then
the limit does not exist.
Example 2.7
Evaluating a Limit That Fails to Exist
Evaluate lim sin(1/x) using a table of values.
x→0
Solution
Table 2.5 lists values for the function sin(1/x) for the given values of x.
x
sin⎛⎝1x⎞⎠
x
sin⎛⎝1x⎞⎠
−0.1
0.544021110889
0.1
−0.544021110889
−0.01
0.50636564111
0.01
−0.50636564111
−0.001
−0.8268795405312
0.001
0.826879540532
−0.0001
0.305614388888
0.0001
−0.305614388888
−0.00001
−0.035748797987
0.00001
0.035748797987
−0.000001
0.349993504187
0.000001
−0.349993504187
Table 2.5
⎛ ⎞
Table of Functional Values for lim sin ⎝1x ⎠
x→0
After examining the table of functional values, we can see that the y-values do not seem to approach any one
single value. It appears the limit does not exist. Before drawing this conclusion, let’s take a more systematic
approach. Take the following sequence of x-values approaching 0:
2 2 2 2 2
2
π , 3π , 5π , 7π , 9π , 11π ,….
The corresponding y-values are
1, −1, 1, −1, 1, −1,….
At this point we can indeed conclude that lim sin(1/x) does not exist. (Mathematicians frequently abbreviate
x→0
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Chapter 2 | Limits
143
“does not exist” as DNE. Thus, we would write lim sin(1/x) DNE.) The graph of f (x) = sin(1/x) is shown
x→0
in Figure 2.17 and it gives a clearer picture of the behavior of sin(1/x) as x approaches 0. You can see that
sin(1/x) oscillates ever more wildly between −1 and 1 as x approaches 0.
Figure 2.17 The graph of f (x) = sin(1/x) oscillates rapidly
between −1 and 1 as x approaches 0.
2.6
Use a table of functional values to evaluate lim
x→2
|xx −−24| ,
2
if possible.
One-Sided Limits
Sometimes indicating that the limit of a function fails to exist at a point does not provide us with enough information
about the behavior of the function at that particular point. To see this, we now revisit the function g(x) = |x − 2|/(x − 2)
introduced at the beginning of the section (see Figure 2.12(b)). As we pick values of x close to 2, g(x) does not approach
a single value, so the limit as x approaches 2 does not exist—that is, lim g(x) DNE. However, this statement alone does
x→2
not give us a complete picture of the behavior of the function around the x-value 2. To provide a more accurate description,
we introduce the idea of a one-sided limit. For all values to the left of 2 (or the negative side of 2), g(x) = −1. Thus, as x
approaches 2 from the left, g(x) approaches −1. Mathematically, we say that the limit as x approaches 2 from the left is −1.
Symbolically, we express this idea as
lim g(x) = −1.
x → 2−
Similarly, as x approaches 2 from the right (or from the positive side), g(x) approaches 1. Symbolically, we express this
idea as
lim g(x) = 1.
x → 2+
We can now present an informal definition of one-sided limits.
Definition
We define two types of one-sided limits.
144
Chapter 2 | Limits
Limit from the left: Let f (x) be a function defined at all values in an open interval of the form z, and let L be a real
number. If the values of the function f (x) approach the real number L as the values of x (where x < a) approach the
number a, then we say that L is the limit of f (x) as x approaches a from the left. Symbolically, we express this idea as
lim
x → a−
f (x) = L.
(2.6)
Limit from the right: Let f (x) be a function defined at all values in an open interval of the form (a, c), and let L be a
real number. If the values of the function f (x) approach the real number L as the values of x (where x > a) approach
the number a, then we say that L is the limit of f (x) as x approaches a from the right. Symbolically, we express this
idea as
lim f (x) = L.
(2.7)
x → a+
Example 2.8
Evaluating One-Sided Limits
⎧x + 1
if x < 2
For the function f (x) = ⎨ 2
, evaluate each of the following limits.
⎩x − 4 if x ≥ 2
a.
b.
lim f (x)
x → 2−
lim f (x)
x → 2+
Solution
We can use tables of functional values again Table 2.6. Observe that for values of x less than 2, we use
f (x) = x + 1 and for values of x greater than 2, we use f (x) = x 2 − 4.
x
f(x) = x + 1
x
f(x) = x2 −4
1.9
2.9
2.1
0.41
1.99
2.99
2.01
0.0401
1.999
2.999
2.001
0.004001
1.9999
2.9999
2.0001
0.00040001
1.99999
2.99999
2.00001
0.0000400001
Table 2.6
⎧x + 1 if
x<2
Table of Functional Values for f (x) = ⎨ 2
⎩x − 4 if x ≥ 2
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Chapter 2 | Limits
145
Based on this table, we can conclude that a.
lim f (x) = 3 and b.
x → 2−
lim f (x) = 0. Therefore, the (two-sided)
x → 2+
limit of f (x) does not exist at x = 2. Figure 2.18 shows a graph of f (x) and reinforces our conclusion about
these limits.
⎧ x + 1 if
x<2
Figure 2.18 The graph of f (x) = ⎨ 2
has a
⎩x − 4 if x ≥ 2
break at x = 2.
2.7
Use a table of functional values to estimate the following limits, if possible.
a.
b.
lim−
x→2
lim
x → 2+
|xx −−24|
2
|xx −−24|
2
Let us now consider the relationship between the limit of a function at a point and the limits from the right and left at that
point. It seems clear that if the limit from the right and the limit from the left have a common value, then that common value
is the limit of the function at that point. Similarly, if the limit from the left and the limit from the right take on different
values, the limit of the function does not exist. These conclusions are summarized in Relating One-Sided and TwoSided Limits.
Theorem 2.2: Relating One-Sided and Two-Sided Limits
Let f (x) be a function defined at all values in an open interval containing a, with the possible exception of a itself,
and let L be a real number. Then,
lim f (x) = L. if and only if lim− f (x) = L and lim f (x) = L.
x→a
x→a
x → a+
146
Chapter 2 | Limits
Infinite Limits
Evaluating the limit of a function at a point or evaluating the limit of a function from the right and left at a point helps us to
characterize the behavior of a function around a given value. As we shall see, we can also describe the behavior of functions
that do not have finite limits.
We now turn our attention to h(x) = 1/(x − 2) 2, the third and final function introduced at the beginning of this section
(see Figure 2.12(c)). From its graph we see that as the values of x approach 2, the values of h(x) = 1/(x − 2) 2 become
larger and larger and, in fact, become infinite. Mathematically, we say that the limit of h(x) as x approaches 2 is positive
infinity. Symbolically, we express this idea as
lim h(x) = +∞.
x→2
More generally, we define infinite limits as follows:
Definition
We define three types of infinite limits.
Infinite limits from the left: Let f (x) be a function defined at all values in an open interval of the form (b, a).
i. If the values of f (x) increase without bound as the values of x (where x < a) approach the number a, then
we say that the limit as x approaches a from the left is positive infinity and we write
lim
x → a−
f (x) = +∞.
(2.8)
ii. If the values of f (x) decrease without bound as the values of x (where x < a) approach the number a, then
we say that the limit as x approaches a from the left is negative infinity and we write
lim
x → a−
f (x) = −∞.
(2.9)
Infinite limits from the right: Let f (x) be a function defined at all values in an open interval of the form (a, c).
i. If the values of f (x) increase without bound as the values of x (where x > a) approach the number a, then
we say that the limit as x approaches a from the left is positive infinity and we write
lim f (x) = +∞.
x → a+
(2.10)
ii. If the values of f (x) decrease without bound as the values of x (where x > a) approach the number a, then
we say that the limit as x approaches a from the left is negative infinity and we write
lim f (x) = −∞.
x → a+
(2.11)
Two-sided infinite limit: Let f (x) be defined for all x ≠ a in an open interval containing a.
i. If the values of f (x) increase without bound as the values of x (where x ≠ a) approach the number a, then
we say that the limit as x approaches a is positive infinity and we write
lim f (x) = +∞.
x→a
(2.12)
ii. If the values of f (x) decrease without bound as the values of x (where x ≠ a) approach the number a, then
we say that the limit as x approaches a is negative infinity and we write
lim f (x) = −∞.
x→a
(2.13)
It is important to understand that when we write statements such as xlim
f (x) = +∞ or xlim
f (x) = −∞ we are
→a
→a
describing the behavior of the function, as we have just defined it. We are not asserting that a limit exists. For the
limit of a function f (x) to exist at a, it must approach a real number L as x approaches a. That said, if, for example,
lim f (x) = +∞, we always write xlim
f (x) = +∞ rather than xlim
f (x) DNE.
→a
→a
x→a
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Chapter 2 | Limits
147
Example 2.9
Recognizing an Infinite Limit
Evaluate each of the following limits, if possible. Use a table of functional values and graph f (x) = 1/x to
confirm your conclusion.
a.
b.
c.
lim 1x
x → 0−
lim 1x
x → 0+
lim 1
x → 0x
Solution
Begin by constructing a table of functional values.
x
1
x
x
1
x
−0.1
−10
0.1
10
−0.01
−100
0.01
100
−0.001
−1000
0.001
1000
−0.0001
−10,000
0.0001
10,000
−0.00001
−100,000
0.00001
100,000
−0.000001
−1,000,000
0.000001
1,000,000
Table 2.7
Table of Functional Values for f (x) = 1x
a. The values of 1/x decrease without bound as x approaches 0 from the left. We conclude that
lim 1x = −∞.
x → 0−
b. The values of 1/x increase without bound as x approaches 0 from the right. We conclude that
lim 1x = +∞.
c. Since
lim 1x = −∞ and
x → 0−
x → 0+
lim 1x = +∞ have different values, we conclude that
x → 0+
lim 1 DNE.
x → 0x
The graph of f (x) = 1/x in Figure 2.19 confirms these conclusions.
148
Chapter 2 | Limits
Figure 2.19 The graph of f (x) = 1/x confirms that the limit
as x approaches 0 does not exist.
2.8
Evaluate each of the following limits, if possible. Use a table of functional values and graph f (x) = 1/x 2
to confirm your conclusion.
a.
b.
c.
lim 1
x → 0− x 2
lim 12
x
x → 0+
lim 1
x → 0 x2
It is useful to point out that functions of the form f (x) = 1/(x − a) n, where n is a positive integer, have infinite limits as x
approaches a from either the left or right (Figure 2.20). These limits are summarized in Infinite Limits from Positive
Integers.
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Chapter 2 | Limits
149
Figure 2.20 The function f (x) = 1/(x − a) n has infinite limits at a.
Theorem 2.3: Infinite Limits from Positive Integers
If n is a positive even integer, then
lim
1
x → a (x − a) n
= +∞.
If n is a positive odd integer, then
lim
1
= +∞
(x − a) n
lim
1
= −∞.
(x − a) n
x → a+
and
x → a−
We should also point out that in the graphs of f (x) = 1/(x − a) n, points on the graph having x-coordinates very near to a
are very close to the vertical line x = a. That is, as x approaches a, the points on the graph of f (x) are closer to the line
x = a. The line x = a is called a vertical asymptote of the graph. We formally define a vertical asymptote as follows:
Definition
Let f (x) be a function. If any of the following conditions hold, then the line x = a is a vertical asymptote of f (x).
lim
x → a−
f (x) = +∞ or −∞
lim f (x) = +∞ or −∞
x → a+
or
(x)
lim
f
= +∞ or −∞
x→a
Example 2.10
150
Chapter 2 | Limits
Finding a Vertical Asymptote
Evaluate each of the following limits using Infinite Limits from Positive Integers. Identify any vertical
asymptotes of the function f (x) = 1/(x + 3) 4.
a.
b.
c.
lim
x → −3−
1
(x + 3) 4
lim
1
(x + 3) 4
x → −3+
1
lim
x → −3 (x + 3) 4
Solution
We can use Infinite Limits from Positive Integers directly.
a.
b.
c.
lim
x → −3−
1
= +∞
(x + 3) 4
lim
1
= +∞
(x + 3) 4
x → −3+
1
lim
x → −3 (x + 3) 4
= +∞
The function f (x) = 1/(x + 3) 4 has a vertical asymptote of x = −3.
2.9
Evaluate each of the following limits. Identify any vertical asymptotes of the function f (x) =
a.
b.
c.
x → 2−
lim
1
(x − 2) 3
lim
1
(x − 2) 3
x → 2+
1 .
(x − 2) 3
1
lim
x → 2 (x − 2) 3
In the next example we put our knowledge of various types of limits to use to analyze the behavior of a function at several
different points.
Example 2.11
Behavior of a Function at Different Points
Use the graph of f (x) in Figure 2.21 to determine each of the following values:
a.
lim
x → −4−
f (x);
lim
x → −4+
f (x); lim f (x); f (−4)
x → −4
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Chapter 2 | Limits
151
b.
lim
x → −2−
c.
x → 1−
d.
x → 3−
f (x);
lim
x → −2+
f (x); lim f (x); f (−2)
x → −2
lim f (x); lim f (x); lim f (x); f (1)
x → 1+
x→1
lim f (x); lim f (x); lim f (x); f (3)
x → 3+
x→3
Figure 2.21 The graph shows f (x).
Solution
Using Infinite Limits from Positive Integers and the graph for reference, we arrive at the following values:
a.
x → −4−
lim
f (x) = 0;
b.
lim
x → −2−
f (x) = 3.;
c.
x → 1−
d.
x → 3−
2.10
lim
f (x) = 0; lim f (x) = 0; f (−4) = 0
lim
f (x) = 3; lim f (x) = 3; f (−2) is undefined
x → −4+
x → −2+
x → −4
x → −2
lim f (x) = 6; lim f (x) = 3; lim f (x) DNE; f (1) = 6
x → 1+
x→1
lim f (x) = −∞; lim f (x) = −∞; lim f (x) = −∞; f (3) is undefined
x → 3+
x→3
Evaluate lim f (x) for f (x) shown here:
x→1
152
Chapter 2 | Limits
Example 2.12
Chapter Opener: Einstein’s Equation
Figure 2.22 (credit: NASA)
In the chapter opener we mentioned briefly how Albert Einstein showed that a limit exists to how fast any object
can travel. Given Einstein’s equation for the mass of a moving object, what is the value of this bound?
Solution
Our starting point is Einstein’s equation for the mass of a moving object,
m=
m0
1−
v2
c2
,
where m 0 is the object’s mass at rest, v is its speed, and c is the speed of light. To see how the mass changes at
high speeds, we can graph the ratio of masses m/m 0 as a function of the ratio of speeds, v/c (Figure 2.23).
Figure 2.23 This graph shows the ratio of masses as a
function of the ratio of speeds in Einstein’s equation for the
mass of a moving object.
We can see that as the ratio of speeds approaches 1—that is, as the speed of the object approaches the speed
of light—the ratio of masses increases without bound. In other words, the function has a vertical asymptote at
v/c = 1. We can try a few values of this ratio to test this idea.
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Chapter 2 | Limits
153
v
c
2
1 − v2
c
m
m0
0.99
0.1411
7.089
0.999
0.0447
22.37
0.9999
0.0141
70.71
Table 2.8
Ratio of Masses and Speeds for a
Moving Object
Thus, according to Table 2.8, if an object with mass 100 kg is traveling at 0.9999c, its mass becomes 7071 kg.
Since no object can have an infinite mass, we conclude that no object can travel at or more than the speed of light.
154
Chapter 2 | Limits
2.2 EXERCISES
For the following exercises, consider the function
2
f (x) = x − 1 .
|x − 1|
30. [T] Complete the following table for the function.
Round your solutions to four decimal places.
f(x)
x
x
f(x)
0.9
a.
1.1
e.
0.99
b.
1.01
f.
0.999
c.
1.001
g.
0.9999
d.
1.0001
h.
31. What do your results in the preceding exercise indicate
about the two-sided limit lim f (x) ? Explain your
x→1
response.
For the following exercises, consider the function
f (x) = (1 + x)
1/x
table to evaluate the limits. Round your solutions to eight
decimal places.
35. [T] lim sin2x
x ; ±0.1, ±0.01, ±0.001, ±.0001
x→0
sin2x
x
x
sin2x
x
x
−0.1
a.
0.1
e.
−0.01
b.
0.01
f.
−0.001
c.
0.001
g.
−0.0001
d.
0.0001
h.
36. [T] lim sin3x
x ±0.1, ±0.01, ±0.001, ±0.0001
x→0
X
sin3x
x
sin3x
x
x
−0.1
a.
0.1
e.
−0.01
b.
0.01
f.
−0.001
c.
0.001
g.
−0.0001
d.
0.0001
h.
.
32.
[T] Make a table showing the values of f for
x = −0.01, −0.001, −0.0001, −0.00001
and
for
x = 0.01, 0.001, 0.0001, 0.00001. Round your solutions
to five decimal places.
x
−0.01
f(x)
a.
x
0.01
f(x)
e.
37. Use the preceding two exercises to conjecture (guess)
−0.001
b.
0.001
f.
−0.0001
c.
0.0001
g.
−0.00001
d.
0.00001
h.
the value of the following limit:
lim sinax
for a, a
x
x→0
positive real value.
33. What does the table of values in the preceding exercise
indicate about the function f (x) = (1 + x) 1/x ?
34. To which mathematical constant does the limit in the
preceding exercise appear to be getting closer?
In the following exercises, use the given values to set up a
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[T] In the following exercises, set up a table of values to
find the indicated limit. Round to eight digits.
Chapter 2 | Limits
lim
38.
155
x2 − 4
x−6
41.
x → 2 x2 +
x2 − 4
x +x−6
x
2
x
x2 − 4
x +x−6
2
lim
z−1
z → 0 z 2 (z + 3)
z −1
z 2 (z + 3)
z
z −1
z 2 (z + 3)
z
1.9
a.
2.1
e.
−0.1
a.
0.1
e.
1.99
b.
2.01
f.
−0.01
b.
0.01
f.
1.999
c.
2.001
g.
−0.001
c.
0.001
g.
1.9999
d.
2.0001
h.
−0.0001
d.
0.0001
h.
39.
40.
lim (1 − 2x)
42.
x→1
x
1 − 2x
x
1 − 2x
0.9
a.
1.1
e.
0.99
b.
1.01
f.
0.999
c.
1.001
g.
0.9999
d.
1.0001
h.
lim
x
t
5
x → 0 1 − e 1/x
5
1 − e 1/x
x
5
1 − e 1/x
−0.1
a.
0.1
e.
−0.01
b.
0.01
f.
−0.001
c.
0.001
g.
−0.0001
d.
0.0001
h.
lim cost
t
t → 0+
43.
cost
t
0.1
a.
0.01
b.
0.001
c.
0.0001
d.
1 − 2x
x → 2 x2 − 4
lim
x
1 − 2x
x2 − 4
x
1 − 2x
x2 − 4
1.9
a.
2.1
e.
1.99
b.
2.01
f.
1.999
c.
2.001
g.
1.9999
d.
2.0001
h.
[T] In the following exercises, set up a table of values
and round to eight significant digits. Based on the table of
values, make a guess about what the limit is. Then, use a
156
Chapter 2 | Limits
calculator to graph the function and determine the limit.
Was the conjecture correct? If not, why does the method of
tables fail?
44.
45.
⎛ ⎞
lim sin ⎝π ⎠
θ
θ→0
θ
⎛ ⎞
sin ⎝π ⎠
θ
θ
⎛ ⎞
sin ⎝π ⎠
θ
−0.1
a.
0.1
e.
−0.01
b.
0.01
f.
−0.001
c.
0.001
g.
−0.0001
d.
0.0001
h.
1 cos ⎛ π ⎞
lim α
⎝α ⎠
+
α→0
a
⎛π ⎞
1
α cos ⎝α ⎠
46.
lim f (x) = 0
x → 10
lim
f (x) = 3
0.1
a.
47.
0.01
b.
48.
0.001
c.
49.
0.0001
d.
In the following exercises, use the following graph of the
function y = f (x) to find the values, if possible. Estimate
x → −2+
lim f (x) = f (−8)
x → −8
lim f (x) = 5
x→6
when necessary.
In the following exercises, consider the graph of the
function y = f (x) shown here. Which of the statements
about y = f (x) are true and which are false? Explain why
a statement is false.
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Chapter 2 | Limits
50.
51.
52.
53.
54.
157
lim f (x)
x → 1−
lim f (x)
x → 1+
lim f (x)
x→1
lim f (x)
x→2
f (1)
In the following exercises, use the graph of the function
y = f (x) shown here to find the values, if possible.
Estimate when necessary.
59.
60.
61.
62.
63.
64.
55.
56.
57.
58.
lim f (x)
x → 0−
lim f (x)
lim
x → −2−
f (x)
lim
f (x)
x → −2+
lim f (x)
x → −2
lim f (x)
x → 2−
lim f (x)
x → 2+
lim f (x)
x→2
In the following exercises, use the graph of the function
y = g(x) shown here to find the values, if possible.
Estimate when necessary.
x → 0+
lim f (x)
x→0
lim f (x)
x→2
In the following exercises, use the graph of the function
y = f (x) shown here to find the values, if possible.
Estimate when necessary.
65.
lim g(x)
x → 0−
158
66.
67.
Chapter 2 | Limits
lim g(x)
x → 0+
lim g(x)
x→0
71.
72.
In the following exercises, use the graph of the function
y = h(x) shown here to find the values, if possible.
73.
Estimate when necessary.
74.
75.
lim f (x)
x → 0−
lim f (x)
x → 0+
lim f (x)
x→0
lim f (x)
x→1
lim f (x)
x→2
In the following exercises, sketch the graph of a function
with the given properties.
76.
lim f (x) = 1,
x→2
lim f (x) = 3,
x → 4−
lim f (x) = 6, x = 4
x → 4+
is not defined.
lim
77.
lim
x → −1+
x→ −∞
69.
70.
lim h(x)
x → 0−
lim h(x)
x → 0+
lim h(x)
x→0
In the following exercises, use the graph of the function
y = f (x) shown here to find the values, if possible.
Estimate when necessary.
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lim
x → −1−
f (x) = −∞,
f (x) = ∞, lim f (x) = f (0), f (0) = 1, x lim
f (x) = −∞
→∞
x→0
lim
78.
68.
f (x) = 0,
x→ −∞
f (x) = 2,
lim f (x) = −∞,
x → 3−
lim f (x) = ∞, x lim
f (x) = 2, f (0) = −1
→∞
3
x → 3+
lim
79.
x→ −∞
f (x) = 2, lim f (x) = −∞,
x → −2
lim f (x) = 2, f (0) = 0
x→∞
80.
lim f (x) = 0,
x→ −∞
f (0) = −1,
lim
x → −1−
f (x) = ∞,
lim f (x) = −∞,
x → 1−
lim
x → −1+
f (x) = −∞,
lim f (x) = ∞, x lim
f (x) = 0
→∞
x → 1+
Chapter 2 | Limits
159
81. Shock waves arise in many physical applications,
ranging from supernovas to detonation waves. A graph of
the density of a shock wave with respect to distance, x, is
shown here. We are mainly interested in the location of the
front of the shock, labeled x SF in the diagram.
a. Evaluate
b. Evaluate
c. Evaluate
lim
ρ(x).
lim
ρ(x).
x → x SF +
x → x SF −
lim ρ(x).
x → x SF
Explain the physical
meanings behind your answers.
82. A track coach uses a camera with a fast shutter to
estimate the position of a runner with respect to time. A
table of the values of position of the athlete versus time is
given here, where x is the position in meters of the runner
and t is time in seconds. What is lim x(t) ? What does it
t→2
mean physically?
t (sec)
x (m)
1.75
4.5
1.95
6.1
1.99
6.42
2.01
6.58
2.05
6.9
2.25
8.5
160
Chapter 2 | Limits
2.3 | The Limit Laws
Learning Objectives
2.3.1
2.3.2
2.3.3
2.3.4
2.3.5
2.3.6
Recognize the basic limit laws.
Use the limit laws to evaluate the limit of a function.
Evaluate the limit of a function by factoring.
Use the limit laws to evaluate the limit of a polynomial or rational function.
Evaluate the limit of a function by factoring or by using conjugates.
Evaluate the limit of a function by using the squeeze theorem.
In the previous section, we evaluated limits by looking at graphs or by constructing a table of values. In this section, we
establish laws for calculating limits and learn how to apply these laws. In the Student Project at the end of this section, you
have the opportunity to apply these limit laws to derive the formula for the area of a circle by adapting a method devised by
the Greek mathematician Archimedes. We begin by restating two useful limit results from the previous section. These two
results, together with the limit laws, serve as a foundation for calculating many limits.
Evaluating Limits with the Limit Laws
The first two limit laws were stated in Two Important Limits and we repeat them here. These basic results, together with
the other limit laws, allow us to evaluate limits of many algebraic functions.
Theorem 2.4: Basic Limit Results
For any real number a and any constant c,
i.
ii.
lim x = a
(2.14)
lim c = c
(2.15)
x→a
x→a
Example 2.13
Evaluating a Basic Limit
Evaluate each of the following limits using Basic Limit Results.
a.
b.
lim x
x→2
lim 5
x→2
Solution
a. The limit of x as x approaches a is a: lim x = 2.
x→2
b. The limit of a constant is that constant: lim 5 = 5.
x→2
We now take a look at the limit laws, the individual properties of limits. The proofs that these laws hold are omitted here.
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Chapter 2 | Limits
161
Theorem 2.5: Limit Laws
Let f (x) and g(x) be defined for all x ≠ a over some open interval containing a. Assume that L and M are real
f (x) = L and xlim
g(x) = M. Let c be a constant. Then, each of the following statements
numbers such that xlim
→a
→a
holds:
⎛
f (x) + xlim
g(x) = L + M
Sum law for limits: xlim
f (x) + g(x)⎞⎠ = xlim
→a
→a
→ a⎝
⎛
f (x) − xlim
g(x) = L − M
f (x) − g(x)⎞⎠ = xlim
Difference law for limits: xlim
→a
→a
→ a⎝
Constant multiple law for limits: xlim
c f (x) = c · xlim
f (x) = cL
→a
→a
⎛
f (x) · xlim
g(x) = L · M
f (x) · g(x)⎞⎠ = xlim
Product law for limits: xlim
→a
→a
→ a⎝
Quotient law for limits: xlim
→a
f (x)
f (x) xlim
= →a
= L for M ≠ 0
M
g(x) xlim
g(x)
→a
⎛
⎞
n
⎛
f (x)⎠ = L n for every positive integer n.
f (x)⎞⎠ n = ⎝xlim
Power law for limits: xlim
→a
→ a⎝
n
n
n
f (x) = xlim
f (x) = L for all L if n is odd and for L ≥ 0 if n is even.
Root law for limits: xlim
→a
→a
We now practice applying these limit laws to evaluate a limit.
Example 2.14
Evaluating a Limit Using Limit Laws
Use the limit laws to evaluate
lim (4x + 2).
x → −3
Solution
Let’s apply the limit laws one step at a time to be sure we understand how they work. We need to keep in mind
the requirement that, at each application of a limit law, the new limits must exist for the limit law to be applied.
lim (4x + 2) = lim 4x + lim 2
x → −3
x → −3
x → −3
Apply the sum law.
= 4 · lim x + lim 2
Apply the constant multiple law.
= 4 · (−3) + 2 = −10.
Apply the basic limit results and simplify.
x → −3
x → −3
Example 2.15
Using Limit Laws Repeatedly
2
3x + 1 .
Use the limit laws to evaluate lim 2x −
x→2
x3 + 4
Solution
162
Chapter 2 | Limits
To find this limit, we need to apply the limit laws several times. Again, we need to keep in mind that as we rewrite
the limit in terms of other limits, each new limit must exist for the limit law to be applied.
2
3x + 1 =
lim 2x −
x→2
x3 + 4
=
lim ⎛⎝2x 2 − 3x + 1⎞⎠
lim ⎛⎝x 3 + 4⎞⎠
x→2
2 · lim x 2 − 3 · lim x + lim 1
x→2
=
2.11
x→2
lim x 3 + lim 4
x→2
2
⎛
=
Apply the quotient law, making sure that. (2) 3 + 4 ≠ 0
x→2
x→2
Apply the sum law and constant multiple law.
x→2
⎞
2 · ⎝ lim x⎠ − 3 · lim x + lim 1
x→2
x→2
x→2
Apply the power law.
⎛
⎞
x + lim 4
⎝xlim
→2 ⎠
x→2
3
2(4) − 3(2) + 1 1
= .
4
(2) 3 + 4
Apply the basic limit laws and simplify.
Use the limit laws to evaluate lim (2x − 1) x + 4. In each step, indicate the limit law applied.
x→6
Limits of Polynomial and Rational Functions
f (x) = f (a). This
By now you have probably noticed that, in each of the previous examples, it has been the case that xlim
→a
is not always true, but it does hold for all polynomials for any choice of a and for all rational functions at all values of a for
which the rational function is defined.
Theorem 2.6: Limits of Polynomial and Rational Functions
Let p(x) and q(x) be polynomial functions. Let a be a real number. Then,
lim p(x) = p(a)
x→a
lim
p(x)
x → a q(x)
=
p(a)
when q(a) ≠ 0.
q(a)
To see that this theorem holds, consider the polynomial p(x) = c n x n + c n − 1 x n − 1 + ⋯ + c 1 x + c 0. By applying the
sum, constant multiple, and power laws, we end up with
⎛
lim p(x) = xlim
c x n + c n − 1 x n − 1 + ⋯ + c 1 x + c 0⎞⎠
→ a⎝ n
x→a
n
x⎞
x⎞ + c n − 1 ⎛⎝xlim
= c n ⎛⎝xlim
→a ⎠
→a ⎠
n−1
x⎞ + lim c
+ ⋯ + c 1 ⎛⎝xlim
→a ⎠ x→a 0
= cn an + cn − 1 an − 1 + ⋯ + c1 a + c0
= p(a).
It now follows from the quotient law that if p(x) and q(x) are polynomials for which q(a) ≠ 0, then
lim
p(x)
x → a q(x)
=
Example 2.16 applies this result.
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p(a)
.
q(a)
Chapter 2 | Limits
163
Example 2.16
Evaluating a Limit of a Rational Function
2
Evaluate the lim 2x − 3x + 1 .
x→3
5x + 4
Solution
2
Since 3 is in the domain of the rational function f (x) = 2x − 3x + 1 , we can calculate the limit by substituting
5x + 4
3 for x into the function. Thus,
2
lim 2x − 3x + 1 = 10 .
19
5x + 4
x→3
2.12
Evaluate
lim ⎛⎝3x 3 − 2x + 7⎞⎠.
x → −2
Additional Limit Evaluation Techniques
As we have seen, we may evaluate easily the limits of polynomials and limits of some (but not all) rational functions by
direct substitution. However, as we saw in the introductory section on limits, it is certainly possible for xlim
f (x) to exist
→a
when f (a) is undefined. The following observation allows us to evaluate many limits of this type:
If for all x ≠ a, f (x) = g(x) over some open interval containing a, then xlim
f (x) = xlim
g(x).
→a
→a
2
To understand this idea better, consider the limit lim x − 1 .
x→1
x−1
The function
2
f (x) = x − 1
x−1
(x − 1)(x + 1)
=
x−1
and the function g(x) = x + 1 are identical for all values of x ≠ 1. The graphs of these two functions are shown in Figure
2.24.
164
Chapter 2 | Limits
Figure 2.24 The graphs of f (x) and g(x) are identical for all x ≠ 1. Their limits at 1 are equal.
We see that
2
(x − 1)(x + 1)
lim x − 1 = lim
x−1
x→1 x−1
x→1
= lim (x + 1)
x→1
= 2.
The limit has the form xlim
→a
f (x)
, where xlim
f (x) = 0 and xlim
g(x) = 0. (In this case, we say that f (x)/g(x) has the
→a
→a
g(x)
indeterminate form 0/0.) The following Problem-Solving Strategy provides a general outline for evaluating limits of this
type.
Problem-Solving Strategy: Calculating a Limit When f(x)/g(x) has the Indeterminate Form 0/0
1. First, we need to make sure that our function has the appropriate form and cannot be evaluated immediately
using the limit laws.
2. We then need to find a function that is equal to h(x) = f (x)/g(x) for all x ≠ a over some interval containing
a. To do this, we may need to try one or more of the following steps:
a. If f (x) and g(x) are polynomials, we should factor each function and cancel out any common factors.
b. If the numerator or denominator contains a difference involving a square root, we should try
multiplying the numerator and denominator by the conjugate of the expression involving the square
root.
c. If f (x)/g(x) is a complex fraction, we begin by simplifying it.
3. Last, we apply the limit laws.
The next examples demonstrate the use of this Problem-Solving Strategy. Example 2.17 illustrates the factor-and-cancel
technique; Example 2.18 shows multiplying by a conjugate. In Example 2.19, we look at simplifying a complex fraction.
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Chapter 2 | Limits
165
Example 2.17
Evaluating a Limit by Factoring and Canceling
x 2 − 3x .
x → 3 2x 2 − 5x − 3
Evaluate lim
Solution
Step 1. The function f (x) =
x 2 − 3x
is undefined for x = 3. In fact, if we substitute 3 into the function
2x − 5x − 3
2
we get 0/0, which is undefined. Factoring and canceling is a good strategy:
x 2 − 3x
lim
x → 3 2x 2 − 5x − 3
Step 2. For all x ≠ 3,
x(x − 3)
= lim
x → 3 (x − 3)(2x + 1)
x 2 − 3x = x . Therefore,
2x − 5x − 3 2x + 1
2
lim
x(x − 3)
x → 3 (x − 3)(2x + 1)
= lim
x
x → 3 2x + 1
.
Step 3. Evaluate using the limit laws:
lim
x
x → 3 2x + 1
2.13
Evaluate
= 3.
7
2
lim x +2 4x + 3 .
x → −3 x − 9
Example 2.18
Evaluating a Limit by Multiplying by a Conjugate
Evaluate
lim
x → −1
x + 2 − 1.
x+1
Solution
Step 1.
x + 2 − 1 has the form 0/0 at −1. Let’s begin by multiplying by
x+1
x + 2 + 1, the conjugate of
x + 2 − 1, on the numerator and denominator:
lim
x → −1
x + 2 − 1 = lim x + 2 − 1 · x + 2 + 1 .
x+1
x → −1 x + 1
x+2+1
Step 2. We then multiply out the numerator. We don’t multiply out the denominator because we are hoping that
the (x + 1) in the denominator cancels out in the end:
= lim
x+1
.
x + 2 + 1⎞⎠
x → −1 (x + 1)⎛⎝
166
Chapter 2 | Limits
Step 3. Then we cancel:
1
.
x+2+1
= lim
x → −1
Step 4. Last, we apply the limit laws:
lim
x → −1
2.14
1
= 1.
x+2+1 2
Evaluate lim x − 1 − 2 .
x→5 x−5
Example 2.19
Evaluating a Limit by Simplifying a Complex Fraction
1 −1
Evaluate lim x + 1 2 .
x→1
x−1
Solution
− 12
has the form 0/0 at 1. We simplify the algebraic fraction by multiplying by
x−1
2(x + 1)/2(x + 1) :
Step 1.
1
x+1
lim
x→1
1 −1
− 12
2(x + 1)
= lim x + 1 2 ·
.
x−1
2(x + 1)
x→1 x−1
1
x+1
Step 2. Next, we multiply through the numerators. Do not multiply the denominators because we want to be able
to cancel the factor (x − 1):
= lim
2 − (x + 1)
x → 1 2(x − 1)(x + 1)
.
Step 3. Then, we simplify the numerator:
−x + 1
= lim
x → 1 2(x − 1)(x + 1)
.
Step 4. Now we factor out −1 from the numerator:
−(x − 1)
= lim
x → 1 2(x − 1)(x + 1)
Step 5. Then, we cancel the common factors of (x − 1):
= lim
−1
x → 1 2(x + 1)
.
Step 6. Last, we evaluate using the limit laws:
lim
−1
x → 1 2(x + 1)
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= − 1.
4
.
Chapter 2 | Limits
2.15
167
Evaluate
lim
x → −3
1
x+2
+1
.
x+3
Example 2.20 does not fall neatly into any of the patterns established in the previous examples. However, with a little
creativity, we can still use these same techniques.
Example 2.20
Evaluating a Limit When the Limit Laws Do Not Apply
⎛
⎞
5
.
Evaluate lim ⎝1x +
x(x − 5) ⎠
x→0
Solution
Both 1/x and 5/x(x − 5) fail to have a limit at zero. Since neither of the two functions has a limit at zero, we
cannot apply the sum law for limits; we must use a different strategy. In this case, we find the limit by performing
addition and then applying one of our previous strategies. Observe that
x−5+5
5
1+
x x(x − 5) = x(x − 5)
x .
=
x(x − 5)
Thus,
⎛
lim 1 +
x → 0⎝ x
5 ⎞ = lim
x
x(x − 5) ⎠
x → 0 x(x − 5)
= lim
1
x → 0x − 5
= − 1.
5
2.16
⎞
⎛ 1
.
− 2 4
⎝
x
−
3
x→3
x − 2x − 3 ⎠
Evaluate lim
Let’s now revisit one-sided limits. Simple modifications in the limit laws allow us to apply them to one-sided limits. For
example, to apply the limit laws to a limit of the form lim− h(x), we require the function h(x) to be defined over an
x→a
open interval of the form (b, a); for a limit of the form
lim h(x), we require the function h(x) to be defined over an
x → a+
open interval of the form (a, c). Example 2.21 illustrates this point.
Example 2.21
Evaluating a One-Sided Limit Using the Limit Laws
Evaluate each of the following limits, if possible.
a.
lim
x → 3−
x−3
168
Chapter 2 | Limits
b.
lim
x → 3+
x−3
Solution
Figure 2.25 illustrates the function f (x) = x − 3 and aids in our understanding of these limits.
Figure 2.25 The graph shows the function f (x) = x − 3.
a. The function f (x) = x − 3 is defined over the interval [3, +∞). Since this function is not defined to
the left of 3, we cannot apply the limit laws to compute
undefined to the left of 3,
lim
x → 3−
lim
x → 3−
x − 3. In fact, since f (x) = x − 3 is
x − 3 does not exist.
b. Since f (x) = x − 3 is defined to the right of 3, the limit laws do apply to
these limit laws we obtain
lim
x → 3+
lim
x → 3+
x − 3. By applying
x − 3 = 0.
In Example 2.22 we look at one-sided limits of a piecewise-defined function and use these limits to draw a conclusion
about a two-sided limit of the same function.
Example 2.22
Evaluating a Two-Sided Limit Using the Limit Laws
⎧4x − 3
For f (x) = ⎨
⎩(x − 3) 2
a.
b.
c.
if x < 2
, evaluate each of the following limits:
if x ≥ 2
lim f (x)
x → 2−
lim f (x)
x → 2+
lim f (x)
x→2
Solution
Figure 2.26 illustrates the function f (x) and aids in our understanding of these limits.
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Chapter 2 | Limits
169
Figure 2.26 This graph shows a function f (x).
a. Since f (x) = 4x − 3 for all x in (−∞, 2), replace f (x) in the limit with 4x − 3 and apply the limit
laws:
lim f (x) = lim− (4x − 3) = 5.
x → 2−
b. Since f (x) = (x − 3)
2
x→2
for all x in (2, +∞), replace f (x) in the limit with (x − 3) 2 and apply the
limit laws:
lim f (x) = lim− (x − 3) 2 = 1.
c. Since
2.17
lim f (x) = 5 and
x → 2−
x → 2+
x→2
lim f (x) = 1, we conclude that lim f (x) does not exist.
x → 2+
x→2
⎧−x − 2 if x < −1
and evaluate lim − f (x).
Graph f (x) = ⎨ 2 if x = −1
x → −1
⎩ x 3 if x > −1
f (x)
We now turn our attention to evaluating a limit of the form xlim
, where xlim
f (x) = K, where K ≠ 0 and
→ a g(x)
→a
lim g(x) = 0. That is, f (x)/g(x) has the form K/0, K ≠ 0 at a.
x→a
Example 2.23
Evaluating a Limit of the Form K/0, K ≠ 0 Using the Limit Laws
Evaluate
lim
x−3 .
x → 2− x 2 − 2x
Solution
Step 1. After substituting in x = 2, we see that this limit has the form −1/0. That is, as x approaches 2 from the
170
Chapter 2 | Limits
left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of
x−3
x(x − 2)
becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:
lim
x − 3 = lim x − 3 .
x → 2− x(x − 2)
x → 2− x 2 − 2x
Step 2. Since x − 2 is the only part of the denominator that is zero when 2 is substituted, we then separate
1/(x − 2) from the rest of the function:
3
= lim− x −
x ·
x→2
3
1
lim x −
x = − 2 and
a limit of +∞:
Step 3.
x → 2−
lim
x → 2−
1 .
x−2
1 = −∞. Therefore, the product of (x − 3)/x and 1/(x − 2) has
x−2
lim
x − 3 = +∞.
x → 2− x 2 − 2x
2.18
Evaluate lim x + 2 2 .
x → 1 (x − 1)
The Squeeze Theorem
The techniques we have developed thus far work very well for algebraic functions, but we are still unable to evaluate limits
of very basic trigonometric functions. The next theorem, called the squeeze theorem, proves very useful for establishing
basic trigonometric limits. This theorem allows us to calculate limits by “squeezing” a function, with a limit at a point a that
is unknown, between two functions having a common known limit at a. Figure 2.27 illustrates this idea.
Figure 2.27 The Squeeze Theorem applies when
f (x) ≤ g(x) ≤ h(x) and lim f (x) = lim h(x).
x→a
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x→a
Chapter 2 | Limits
171
Theorem 2.7: The Squeeze Theorem
Let f (x), g(x), and h(x) be defined for all x ≠ a over an open interval containing a. If
f (x) ≤ g(x) ≤ h(x)
for all x ≠ a in an open interval containing a and
lim f (x) = L = xlim
h(x)
→a
x→a
where L is a real number, then xlim
g(x) = L.
→a
Example 2.24
Applying the Squeeze Theorem
Apply the squeeze theorem to evaluate lim x cos x.
x→0
Solution
Because −1 ≤ cos x ≤ 1 for all x, we have −x ≤ x cos x ≤ x for x ≥ 0 and −x ≥ xcosx ≥ x for x ≤ 0 (if
x is negative the direction of the inequalities changes when we multiply). Since lim (−x) = 0 = lim x, from
x→0
x→0
the squeeze theorem, we obtain lim x cos x = 0. The graphs of f (x) = − x, g(x) = x cos x, and h(x) = x are
x→0
shown in Figure 2.28.
Figure 2.28 The graphs of f (x), g(x), and h(x) are shown
around the point x = 0.
2.19
Use the squeeze theorem to evaluate lim x 2 sin 1x .
x→0
We now use the squeeze theorem to tackle several very important limits. Although this discussion is somewhat lengthy,
these limits prove invaluable for the development of the material in both the next section and the next chapter. The first of
these limits is lim sinθ. Consider the unit circle shown in Figure 2.29. In the figure, we see that sinθ is the y-coordinate
θ→0
on the unit circle and it corresponds to the line segment shown in blue. The radian measure of angle θ is the length of the
arc it subtends on the unit circle. Therefore, we see that for 0 < θ < π , 0 < sinθ < θ.
2
172
Chapter 2 | Limits
Figure 2.29 The sine function is shown as a line on the unit
circle.
Because
lim 0 = 0 and
θ → 0+
lim θ = 0, by using the squeeze theorem we conclude that
x → 0+
lim sinθ = 0.
θ → 0+
lim sinθ = 0 as well, observe that for − π < θ < 0, 0 < −θ < π and hence, 0 < sin(−θ) < −θ.
2
2
Consequently, 0 < − sinθ < −θ. It follows that 0 > sinθ > θ. An application of the squeeze theorem produces the
desired limit. Thus, since lim sinθ = 0 and lim− sinθ = 0,
To see that
θ → 0−
θ → 0+
θ→0
lim sinθ = 0.
(2.16)
θ→0
Next, using the identity cosθ = 1 − sin 2 θ for − π < θ < π , we see that
2
2
(2.17)
lim cosθ = lim 1 − sin 2 θ = 1.
θ→0
θ→0
We now take a look at a limit that plays an important role in later chapters—namely, lim sinθ . To evaluate this limit,
θ→0
θ
we use the unit circle in Figure 2.30. Notice that this figure adds one additional triangle to Figure 2.30. We see that the
length of the side opposite angle θ in this new triangle is tanθ. Thus, we see that for 0 < θ < π , sinθ < θ < tanθ.
2
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Chapter 2 | Limits
173
Figure 2.30 The sine and tangent functions are shown as lines
on the unit circle.
By dividing by sinθ in all parts of the inequality, we obtain
1< θ < 1 .
sinθ cosθ
Equivalently, we have
1 > sinθ > cosθ.
θ
Since
lim 1 = 1 = lim cosθ, we conclude that
θ → 0+
θ → 0+
in demonstrating that
lim sinθ = 1. By applying a manipulation similar to that used
θ
θ → 0+
lim sinθ = 0, we can show that
θ → 0−
lim sinθ = 1. Thus,
θ
θ → 0−
lim sinθ = 1.
θ
θ→0
(2.18)
In Example 2.25 we use this limit to establish lim 1 − cosθ = 0. This limit also proves useful in later chapters.
θ→0
θ
Example 2.25
Evaluating an Important Trigonometric Limit
Evaluate lim 1 − cosθ .
θ→0
θ
Solution
In the first step, we multiply by the conjugate so that we can use a trigonometric identity to convert the cosine in
the numerator to a sine:
174
Chapter 2 | Limits
lim 1 − cosθ = lim 1 − cosθ · 1 + cosθ
θ
θ
1 + cosθ
θ→0
θ→0
2
= lim 1 − cos θ
θ → 0 θ(1 + cosθ)
= lim
sin 2 θ
θ → 0 θ(1 + cosθ)
= lim sinθ · sinθ
1 + cosθ
θ→0 θ
0
= 1 · = 0.
2
Therefore,
lim 1 − cosθ = 0.
θ
θ→0
2.20
Evaluate lim 1 − cosθ .
θ → 0 sinθ
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(2.19)
Chapter 2 | Limits
175
Deriving the Formula for the Area of a Circle
Some of the geometric formulas we take for granted today were first derived by methods that anticipate some of the
methods of calculus. The Greek mathematician Archimedes (ca. 287−212; BCE) was particularly inventive, using
polygons inscribed within circles to approximate the area of the circle as the number of sides of the polygon increased.
He never came up with the idea of a limit, but we can use this idea to see what his geometric constructions could have
predicted about the limit.
We can estimate the area of a circle by computing the area of an inscribed regular polygon. Think of the regular
polygon as being made up of n triangles. By taking the limit as the vertex angle of these triangles goes to zero, you can
obtain the area of the circle. To see this, carry out the following steps:
1. Express the height h and the base b of the isosceles triangle in Figure 2.31 in terms of θ and r.
Figure 2.31
2. Using the expressions that you obtained in step 1, express the area of the isosceles triangle in terms of θ and r.
(Substitute (1 / 2)sinθ for sin(θ / 2)cos(θ / 2) in your expression.)
3. If an n-sided regular polygon is inscribed in a circle of radius r, find a relationship between θ and n. Solve this
for n. Keep in mind there are 2π radians in a circle. (Use radians, not degrees.)
4. Find an expression for the area of the n-sided polygon in terms of r and θ.
5. To find a formula for the area of the circle, find the limit of the expression in step 4 as θ goes to zero. (Hint:
lim
θ→0
(sinθ)
= 1).
θ
The technique of estimating areas of regions by using polygons is revisited in Introduction to Integration.
176
Chapter 2 | Limits
2.3 EXERCISES
In the following exercises, use the limit laws to evaluate
each limit. Justify each step by indicating the appropriate
limit law(s).
83.
84.
85.
86.
lim ⎛⎝4x 2 − 2x + 3⎞⎠
x→0
3
2
lim x + 3x + 5
4 − 7x
x→1
lim
x → −2
x 2 − 6x + 3
lim (9x + 1) 2
x → −1
In the following exercises, use direct substitution to
evaluate each limit.
87.
88.
89.
90.
91.
92.
h→0
1
a+h
h
− 1a
, where a is a real-valued constant
lim sinθ
θ → π tanθ
100.
3
lim x 2 − 1
x → 1x − 1
101.
2
lim 2x + 3x − 2
2x − 1
x → 1/2
102.
x+4−1
x+3
lim
x → −3
lim x 2
lim ⎛⎝4x 2 − 1⎞⎠
103.
x→7
x → −2
lim
1
x → 0 1 + sin x
lim e 2x − x
104.
2
x→2
lim 2 − 7x
x+6
x→1
lim x − 2
(1 + h) 2 − 1
h
h→0
97.
lim t − 9
t→9 t−3
2x 2 + 7x − 4
x2 + x − 2
2
2
lim 2x 2 + 7x − 4
+
x +x−2
x→1
In
the
following
exercises,
assume
that
lim f (x) = 4, lim g(x) = 9, and lim h(x) = 6. Use
x→6
x→6
x→6
these three facts and the limit laws to evaluate each limit.
107.
108.
lim 2 f (x)g(x)
x→6
lim
g(x) − 1
f (x)
x→6
⎛
x → 6⎝
110.
h(x)⎞⎠ 3
x→6 2
111.
112.
⎞
lim f (x) + 1 g(x)⎠
3
109.
lim
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x → −2+
106.
x → 6 2x − 12
96.
lim
lim 2x + 7x − 4
x → 1− x 2 + x − 2
x → 2 x 2 − 2x
lim 3x − 18
2
lim 2x + 7x − 4
x → −2− x 2 + x − 2
105.
lim lne 3x
x→3
2
93. lim x − 16
x→4 x−4
95.
99.
lim
In the following exercises, use direct substitution to obtain
an undefined expression. Then, use the method of
Example 2.23 to simplify the function to help determine
the limit.
In the following exercises, use direct substitution to show
that each limit leads to the indeterminate form 0/0. Then,
evaluate the limit.
94.
98.
lim
⎛
⎝
lim g(x) − f (x)
x→6
lim x · h(x)
x→6
Chapter 2 | Limits
113.
114.
177
lim ⎡⎣(x + 1) · f (x)⎤⎦
x→6
lim ⎛⎝ f (x) · g(x) − h(x)⎞⎠
x→6
[T] In the following exercises, use a calculator to draw
the graph of each piecewise-defined function and study the
graph to evaluate the given limits.
⎧x 2,
x≤3
f (x) = ⎨
⎩x + 4, x > 3
a.
lim− f (x)
115.
x→3
b.
lim f (x)
x → 3+
⎧x 3 − 1,
116. g(x) = ⎨
⎩1,
a.
b.
x≤0
x>0
lim g(x)
x → 0−
lim g(x)
x → 0+
⎧x 2 − 2x + 1,
117. h(x) = ⎨
⎩3 −
a.
b.
x,
x<2
x≥2
lim h(x)
x → 2−
lim h(x)
x → 2+
In the following exercises, use the following graphs and the
limit laws to evaluate each limit.
118.
119.
120.
121.
122.
123.
124.
lim
⎛
⎝
f (x) + g(x)⎞⎠
lim
⎛
⎝
f (x) − 3g(x)⎞⎠
x → −3+
x → −3−
lim
x→0
f (x)g(x)
3
lim
x → −5
2 + g(x)
f (x)
lim ⎛⎝ f (x)⎞⎠ 2
x→1
lim f (x) − g(x)
x→1
lim ⎛⎝x · g(x)⎞⎠
x → −7
178
Chapter 2 | Limits
125.
lim ⎡⎣x · f (x) + 2 · g(x)⎤⎦
x → −9
For the following problems, evaluate the limit using the
squeeze theorem. Use a calculator to graph the functions
f (x), g(x), and h(x) when possible.
126.
[T]
True
or
False?
2x − 1 ≤ g(x) ≤ x 2 − 2x + 3, then lim g(x) = 0.
If
x→2
⎛ ⎞
127. [T] lim θ 2 cos ⎝1 ⎠
θ
θ→0
128.
⎧0,
x rational
lim f (x), where f (x) = ⎨ 2
x→0
⎩x , x irrrational
129. [T] In physics, the magnitude of an electric field
generated by a point charge at a distance r in vacuum
is governed by Coulomb’s law: E(r) =
q
, where
4πε 0 r 2
E represents the magnitude of the electric field, q is the
charge of the particle, r is the distance between the particle
and where the strength of the field is measured, and
1
4πε 0
is Coulomb’s constant: 8.988 × 10 9 N · m 2 /C 2.
a. Use a graphing calculator to graph E(r) given that
the charge of the particle is q = 10 −10.
b. Evaluate
lim E(r).
r → 0+
What is the physical
meaning of this quantity? Is it physically relevant?
Why are you evaluating from the right?
130. [T] The density of an object is given by its mass
divided by its volume: ρ = m/V.
a. Use a calculator to plot the volume as a function of
density ⎛⎝V = m/ρ⎞⎠, assuming you are examining
something of mass 8 kg ( m = 8).
b. Evaluate
lim V(ρ) and explain the physical
ρ → 0+
meaning.
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Chapter 2 | Limits
179
2.4 | Continuity
Learning Objectives
2.4.1
2.4.2
2.4.3
2.4.4
2.4.5
Explain the three conditions for continuity at a point.
Describe three kinds of discontinuities.
Define continuity on an interval.
State the theorem for limits of composite functions.
Provide an example of the intermediate value theorem.
Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such
functions are called continuous. Other functions have points at which a break in the graph occurs, but satisfy this property
over intervals contained in their domains. They are continuous on these intervals and are said to have a discontinuity at a
point where a break occurs.
We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively,
a function is continuous at a particular point if there is no break in its graph at that point.
Continuity at a Point
Before we look at a formal definition of what it means for a function to be continuous at a point, let’s consider various
functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions
that prevent such failures.
Our first function of interest is shown in Figure 2.32. We see that the graph of f (x) has a hole at a. In fact, f (a) is
undefined. At the very least, for f (x) to be continuous at a, we need the following condition:
i. f (a) is defined
Figure 2.32 The function f (x) is not continuous at a
because f (a) is undefined.
However, as we see in Figure 2.33, this condition alone is insufficient to guarantee continuity at the point a. Although
f (a) is defined, the function has a gap at a. In this example, the gap exists because xlim
f (x) does not exist. We must add
→a
another condition for continuity at a—namely,
ii. xlim
f (x) exists.
→a
180
Chapter 2 | Limits
Figure 2.33 The function f (x) is not continuous at a
because lim f (x) does not exist.
x→a
However, as we see in Figure 2.34, these two conditions by themselves do not guarantee continuity at a point. The function
in this figure satisfies both of our first two conditions, but is still not continuous at a. We must add a third condition to our
list:
iii. xlim
f (x) = f (a).
→a
Figure 2.34 The function f (x) is not continuous at a
because lim f (x) ≠ f (a).
x→a
Now we put our list of conditions together and form a definition of continuity at a point.
Definition
A function f (x) is continuous at a point a if and only if the following three conditions are satisfied:
i.
ii.
iii.
f (a) is defined
lim f (x) exists
x→a
lim f (x) = f (a)
x→a
A function is discontinuous at a point a if it fails to be continuous at a.
The following procedure can be used to analyze the continuity of a function at a point using this definition.
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Chapter 2 | Limits
181
Problem-Solving Strategy: Determining Continuity at a Point
1. Check to see if f (a) is defined. If f (a) is undefined, we need go no further. The function is not continuous
at a. If f (a) is defined, continue to step 2.
2. Compute xlim
f (x). In some cases, we may need to do this by first computing lim− f (x) and lim f (x).
→a
x→a
x → a+
If xlim
f (x) does not exist (that is, it is not a real number), then the function is not continuous at a and the
→a
problem is solved. If xlim
f (x) exists, then continue to step 3.
→a
3. Compare f (a) and xlim
f (x). If xlim
f (x) ≠ f (a),
→a
→a
lim f (x) = f (a), then the function is continuous at a.
x→a
then the function is not continuous at a. If
The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given
point. These examples illustrate situations in which each of the conditions for continuity in the definition succeed or fail.
Example 2.26
Determining Continuity at a Point, Condition 1
Using the definition, determine whether the function f (x) = (x 2 − 4)/(x − 2) is continuous at x = 2. Justify
the conclusion.
Solution
Let’s begin by trying to calculate f (2). We can see that f (2) = 0/0,
which is undefined. Therefore,
2
f (x) = x − 4 is discontinuous at 2 because f (2) is undefined. The graph of f (x) is shown in Figure 2.35.
x−2
Figure 2.35 The function f (x) is discontinuous at 2 because
f (2) is undefined.
182
Chapter 2 | Limits
Example 2.27
Determining Continuity at a Point, Condition 2
⎧
2
Using the definition, determine whether the function f (x) = ⎨−x + 4 if x ≤ 3 is continuous at x = 3. Justify
⎩4x − 8
if x > 3
the conclusion.
Solution
Let’s begin by trying to calculate f (3).
f (3) = − (3 2) + 4 = −5.
Thus, f (3) is defined. Next, we calculate
lim f (x):
lim f (x). To do this, we must compute
x→3
lim f (x) and
x → 3−
x → 3+
lim f (x) = − (3 2) + 4 = −5
x → 3−
and
lim f (x) = 4(3) − 8 = 4.
x → 3+
Therefore, lim f (x) does not exist. Thus, f (x) is not continuous at 3. The graph of f (x) is shown in Figure
x→3
2.36.
Figure 2.36 The function f (x) is not continuous at 3
because lim f (x) does not exist.
x→3
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Chapter 2 | Limits
183
Example 2.28
Determining Continuity at a Point, Condition 3
⎧sin x
Using the definition, determine whether the function f (x) = ⎨ x
⎩
if x ≠ 0
1 if x = 0
is continuous at x = 0.
Solution
First, observe that
f (0) = 1.
Next,
x = 1.
lim f (x) = lim sin
x→0 x
x→0
Last, compare f (0) and lim f (x). We see that
x→1
f (0) = 1 = lim f (x).
x→0
Since all three of the conditions in the definition of continuity are satisfied, f (x) is continuous at x = 0.
⎧2x + 1 if x < 1
Using the definition, determine whether the function f (x) = ⎨
2 if x = 1 is continuous at x = 1.
⎩−x + 4 if x > 1
If the function is not continuous at 1, indicate the condition for continuity at a point that fails to hold.
2.21
By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can
state the following theorem.
Theorem 2.8: Continuity of Polynomials and Rational Functions
Polynomials and rational functions are continuous at every point in their domains.
Proof
p(x) = p(a) for every polynomial p(x) and
Previously, we showed that if p(x) and q(x) are polynomials, xlim
→a
lim
p(x)
x → a q(x)
=
p(a)
as long as q(a) ≠ 0. Therefore, polynomials and rational functions are continuous on their domains.
q(a)
□
We now apply Continuity of Polynomials and Rational Functions to determine the points at which a given rational
function is continuous.
Example 2.29
Continuity of a Rational Function
184
Chapter 2 | Limits
For what values of x is f (x) = x + 1 continuous?
x−5
Solution
The rational function f (x) = x + 1 is continuous for every value of x except x = 5.
x−5
2.22
For what values of x is f (x) = 3x 4 − 4x 2 continuous?
Types of Discontinuities
As we have seen in Example 2.26 and Example 2.27, discontinuities take on several different appearances. We
classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump
discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump
discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite
discontinuity is a discontinuity located at a vertical asymptote. Figure 2.37 illustrates the differences in these types of
discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in
mind that not all discontinuities fit neatly into these categories.
Figure 2.37 Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.
These three discontinuities are formally defined as follows:
Definition
If f (x) is discontinuous at a, then
1.
f has a removable discontinuity at a if xlim
f (x) exists. (Note: When we state that xlim
f (x) exists, we
→a
→a
mean that xlim
f (x) = L, where L is a real number.)
→a
2.
f has a jump discontinuity at a if
(Note: When we state that
lim
x → a−
lim f (x) and
x → a−
f (x) and
lim f (x) both exist, but
x → a+
lim
x → a−
f has an infinite discontinuity at a if
x → a+
lim f (x) both exist, we mean that both are real-valued and that
x → a+
neither take on the values ±∞.)
3.
f (x) ≠ lim f (x).
lim
x → a−
f (x) = ±∞ or
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lim f (x) = ±∞.
x → a+
Chapter 2 | Limits
185
Example 2.30
Classifying a Discontinuity
2
In Example 2.26, we showed that f (x) = x − 4 is discontinuous at x = 2. Classify this discontinuity as
x−2
removable, jump, or infinite.
Solution
To classify the discontinuity at 2 we must evaluate lim f (x):
x→2
2
lim f (x) = lim x − 4
x→2
x→2 x−2
(x − 2)(x + 2)
= lim
x−2
x→2
= lim (x + 2)
x→2
= 4.
Since f is discontinuous at 2 and lim f (x) exists, f has a removable discontinuity at x = 2.
x→2
Example 2.31
Classifying a Discontinuity
⎧
2
In Example 2.27, we showed that f (x) = ⎨−x + 4 if x ≤ 3 is discontinuous at x = 3. Classify this
⎩
4x − 8 if x > 3
discontinuity as removable, jump, or infinite.
Solution
Earlier, we showed that f is discontinuous at 3 because
lim f (x) = −5 and
x → 3−
lim f (x) does not exist. However, since
x→3
lim f (x) = 4 both exist, we conclude that the function has a jump discontinuity at 3.
x → 3−
Example 2.32
Classifying a Discontinuity
Determine whether f (x) = x + 2 is continuous at −1. If the function is discontinuous at −1, classify the
x+1
discontinuity as removable, jump, or infinite.
Solution
The function value f (−1) is undefined. Therefore, the function is not continuous at −1. To determine the type of
186
Chapter 2 | Limits
discontinuity, we must determine the limit at −1. We see that
lim
x → −1−
x + 2 = −∞ and
x+1
lim
x → −1+
x + 2 = +∞.
x+1
Therefore, the function has an infinite discontinuity at −1.
⎧ 2
For f (x) = ⎨x if x ≠ 1, decide whether f is continuous at 1. If f is not continuous at 1, classify the
⎩ 3 if x = 1
discontinuity as removable, jump, or infinite.
2.23
Continuity over an Interval
Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As
we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive idea that a function is
continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting
the pencil from the paper. In preparation for defining continuity on an interval, we begin by looking at the definition of what
it means for a function to be continuous from the right at a point and continuous from the left at a point.
Continuity from the Right and from the Left
A function f (x) is said to be continuous from the right at a if
A function f (x) is said to be continuous from the left at a if
lim f (x) = f (a).
x → a+
lim f (x) = f (a).
x → a−
A function is continuous over an open interval if it is continuous at every point in the interval. A function f (x) is continuous
over a closed interval of the form ⎡⎣a, b⎤⎦ if it is continuous at every point in (a, b) and is continuous from the right at a
and is continuous from the left at b. Analogously, a function f (x) is continuous over an interval of the form (a, b⎤⎦ if it is
continuous over (a, b) and is continuous from the left at b. Continuity over other types of intervals are defined in a similar
fashion.
Requiring that
lim f (x) = f (a) and
x → a+
lim f (x) = f (b) ensures that we can trace the graph of the function from the
x → b−
point ⎛⎝a, f (a)⎞⎠ to the point ⎛⎝b, f (b)⎞⎠ without lifting the pencil. If, for example,
lim f (x) ≠ f (a), we would need to lift
x → a+
our pencil to jump from f (a) to the graph of the rest of the function over (a, b⎤⎦.
Example 2.33
Continuity on an Interval
State the interval(s) over which the function f (x) = x2 − 1
x + 2x
is continuous.
Solution
Since f (x) = x2 − 1
is a rational function, it is continuous at every point in its domain. The domain of
x + 2x
f (x) is the set (−∞, −2) ∪ (−2, 0) ∪ (0, +∞). Thus, f (x) is continuous over each of the intervals
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Chapter 2 | Limits
187
(−∞, −2), (−2, 0), and (0, +∞).
Example 2.34
Continuity over an Interval
State the interval(s) over which the function f (x) = 4 − x 2 is continuous.
Solution
From the limit laws, we know that xlim
4 − x 2 = 4 − a 2 for all values of a in (−2, 2). We also know that
→a
lim
x → −2+
4 − x 2 = 0 exists and
lim
x → 2−
4 − x 2 = 0 exists. Therefore, f (x) is continuous over the interval
[−2, 2].
2.24
State the interval(s) over which the function f (x) = x + 3 is continuous.
The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem
ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.
Theorem 2.9: Composite Function Theorem
g(x) = L, then
If f (x) is continuous at L and xlim
→a
g(x)⎞⎠ = f (L).
lim f ⎛⎝g(x)⎞⎠ = f ⎛⎝xlim
→a
x→a
Before we move on to Example 2.35, recall that earlier, in the section on limit laws, we showed lim cos x = 1 = cos(0).
x→0
Consequently, we know that f (x) = cos x is continuous at 0. In Example 2.35 we see how to combine this result with the
composite function theorem.
Example 2.35
Limit of a Composite Cosine Function
Evaluate
⎛
Solution
⎞
lim cos ⎝x − π ⎠.
2
x → π/2
188
Chapter 2 | Limits
The given function is a composite of cos x and x − π . Since
2
we may apply the composite function theorem. Thus,
⎛
⎞
⎛
lim
⎛
⎛
⎞
x − π ⎠ = 0 and cos x is continuous at 0,
2
x → π/2⎝
⎞⎞
lim cos ⎝x − π ⎠ = cos ⎝ lim ⎝x − π ⎠⎠ = cos(0) = 1.
2
2
x → π/2
x → π/2
2.25
Evaluate xlim
sin(x − π).
→π
The proof of the next theorem uses the composite function theorem as well as the continuity of f (x) = sin x and
g(x) = cos x at the point 0 to show that trigonometric functions are continuous over their entire domains.
Theorem 2.10: Continuity of Trigonometric Functions
Trigonometric functions are continuous over their entire domains.
Proof
cos x = cosa
We begin by demonstrating that cos x is continuous at every real number. To do this, we must show that xlim
→a
for all values of a.
lim cos x = xlim
cos((x − a) + a)
→a
x→a
⎛
⎝
rewrite x = x − a + a
⎞
⎠
= xlim
cos(x − a)cosa − sin(x − a)sina
→a
(x − a)⎞⎠ sina
(x − a)⎞⎠ cosa − sin ⎛⎝xlim
= cos ⎛⎝xlim
→a
→a
= cos(0)cosa − sin(0)sina
= 1 · cosa − 0 · sina = cosa.
apply the identity for the cosine of the sum of two angles
lim (x − a) = 0, and sin x and cos x are continuous at 0
x→a
evaluate cos(0) and sin(0) and simplify
The proof that sin x is continuous at every real number is analogous. Because the remaining trigonometric functions may
be expressed in terms of sin x and cos x, their continuity follows from the quotient limit law.
□
As you can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions.
As we continue our study of calculus, we revisit this theorem many times.
The Intermediate Value Theorem
Functions that are continuous over intervals of the form ⎡⎣a, b⎤⎦, where a and b are real numbers, exhibit many useful
properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The
first of these theorems is the Intermediate Value Theorem.
Theorem 2.11: The Intermediate Value Theorem
Let f be continuous over a closed, bounded interval ⎡⎣a, b⎤⎦. If z is any real number between f (a) and f (b), then there
is a number c in ⎡⎣a, b⎤⎦ satisfying f (c) = z in Figure 2.38.
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Chapter 2 | Limits
189
Figure 2.38 There is a number c ∈ ⎡⎣a, b⎤⎦ that satisfies
f (c) = z.
Example 2.36
Application of the Intermediate Value Theorem
Show that f (x) = x − cos x has at least one zero.
Solution
Since f (x) = x − cos x is continuous over (−∞, +∞), it is continuous over any closed interval of the form
⎡
⎣
a, b⎤⎦. If you can find an interval ⎡⎣a, b⎤⎦ such that f (a) and f (b) have opposite signs, you can use the
Intermediate Value Theorem to conclude there must be a real number c in (a, b) that satisfies f (c) = 0. Note
that
f (0) = 0 − cos(0) = −1 < 0
and
⎛ ⎞
f ⎝π ⎠ = π − cos π = π > 0.
2
2 2
2
Using the Intermediate Value Theorem, we can see that there must be a real number c in [0, π/2] that satisfies
f (c) = 0. Therefore, f (x) = x − cos x has at least one zero.
Example 2.37
When Can You Apply the Intermediate Value Theorem?
If f (x) is continuous over [0, 2], f (0) > 0 and f (2) > 0, can we use the Intermediate Value Theorem to
conclude that f (x) has no zeros in the interval ⎡⎣0, 2]? Explain.
190
Chapter 2 | Limits
Solution
No. The Intermediate Value Theorem only allows us to conclude that we can find a value between f (0) and
f (2); it doesn’t allow us to conclude that we can’t find other values. To see this more clearly, consider the
function f (x) = (x − 1) 2. It satisfies f (0) = 1 > 0, f (2) = 1 > 0, and f (1) = 0.
Example 2.38
When Can You Apply the Intermediate Value Theorem?
For f (x) = 1/x, f (−1) = −1 < 0 and f (1) = 1 > 0. Can we conclude that f (x) has a zero in the interval
[−1, 1] ?
Solution
No. The function is not continuous over [−1, 1]. The Intermediate Value Theorem does not apply here.
2.26
Show that f (x) = x 3 − x 2 − 3x + 1 has a zero over the interval [0, 1].
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Chapter 2 | Limits
191
2.4 EXERCISES
For the following exercises, determine the point(s), if any,
at which each function is discontinuous. Classify any
discontinuity as jump, removable, infinite, or other.
131.
f (x) = 1
x
⎧
sinθ, 0 ≤ θ < π
2
146. f (θ) = ⎨
π ≤θ≤π
cos(θ
+
k),
⎩
2
147.
2
x +1
⎧ x 2 + 3x + 2
f (x) = ⎨
⎩
x+2
, x≠ −2
k, x = −2
132.
f (x) =
133.
f (x) = 2 x
x −x
148.
⎧ e kx, 0 ≤ x < 4
f (x) = ⎨
⎩x + 3, 4 ≤ x ≤ 8
134. g(t) = t −1 + 1
149.
⎧ kx, 0 ≤ x ≤ 3
f (x) = ⎨
⎩x + 1, 3 < x ≤ 10
2
5
ex − 2
135.
f (x) =
136.
f (x) = |x − 2|
x−2
In the following exercises, use the Intermediate Value
Theorem (IVT).
150.
137. H(x) = tan2x
138.
f (t) =
[0, 4],
does not contradict the IVT.
For the following exercises, decide if the function
continuous at the given point. If it is discontinuous, what
type of discontinuity is it?
2
139. 2x − 5x + 3 at x = 1
x−1
140. h(θ) = sinθ − cosθ at θ = π
tanθ
141. g(u) =
if u ≠ 1
2 , at u = 1
2
if u = 1
2
f (y) =
sin(πy)
, at y = 1
tan(πy)
143.
f (x) =
⎧ 2
x
⎨x − e
⎩x − 1
144.
⎧x sin(x) if x ≤ π
f (x) = ⎨
, at x = π
⎩x tan(x) if x > π
142.
if x < 0, at x = 0
if x ≥ 0
In the following exercises, find the value(s) of k that makes
each function continuous over the given interval.
145.
x≤2
Over the interval
x>2
there is no value of x such that h(x) = 10,
⎩5 + 4x,
although h(0) < 10 and h(4) > 10. Explain why this
t+3
t 2 + 5t + 6
⎧6u 2 + u − 2
⎪
⎨ 2u − 1
⎪7
⎩2
⎧3x 2 − 4,
Let h(x) = ⎨
⎧3x + 2, x < k
f (x) = ⎨
⎩2x − 3, k ≤ x ≤ 8
151. A particle moving along a line has at each time t
a position function s(t), which is continuous. Assume
s(2) = 5 and s(5) = 2. Another particle moves such that
its position is given by h(t) = s(t) − t. Explain why there
must be a value c for 2 < c < 5 such that h(c) = 0.
152. [T] Use the statement “The cosine of t is equal to t
cubed.”
a. Write a mathematical equation of the statement.
b. Prove that the equation in part a. has at least one
real solution.
c. Use a calculator to find an interval of length 0.01
that contains a solution.
153. Apply the IVT to determine whether 2 x = x 3 has
a solution in one of the intervals ⎡⎣1.25, 1.375⎤⎦ or
⎡
⎣
1.375, 1.5⎤⎦. Briefly explain your response for each
interval.
192
Chapter 2 | Limits
154. Consider the graph of the function y = f (x) shown
158.
Sketch the graph of the function y = f (x) with
in the following graph.
properties i. through iv.
i. The domain of f is ⎡⎣0, 5⎤⎦.
ii.
lim f (x) and
x → 1+
lim f (x) exist and are equal.
x → 1−
iii.
f (x) is left continuous but not continuous at
iv.
x = 2, and right continuous but not continuous at
x = 3.
f (x) has a removable discontinuity at x = 1, a
jump discontinuity at x = 2, and the following
limits
a. Find all values for which the function is
discontinuous.
b. For each value in part a., state why the formal
definition of continuity does not apply.
c. Classify each discontinuity as either jump,
removable, or infinite.
⎧3x,
x>1
a. Sketch the graph of f.
b. Is it possible to find a value k such that f (1) = k,
f (x)
continuous for all real
numbers? Briefly explain.
continuous for all real numbers? Briefly explain.
Sketch the graph of the function y = f (x) with
properties i. through vii.
i. The domain of f is (−∞, +∞).
f (x) =
lim
x → −3−
v.
f (−3) = 3
lim
x → −3+
f (x) = 2
all x. For each description, sketch a graph with the indicated
property.
x = 3.
lim f (x) = −∞ and
x→ −∞
lim f (x) = 4
lim
x→ +∞
lim f (x) = −1 and
x → −1
x→2
160. Discontinuous at x = 2 but continuous elsewhere
with lim f (x) = 1
2
Determine whether each of the given statements is true.
Justify your response with an explanation or
counterexample.
161.
f (t) =
2
is continuous everywhere.
e t − e −t
162. If the left- and right-hand limits of f (x) as x → a
exist and are equal, then f cannot be discontinuous at
163. If a function is not continuous at a point, then it is not
defined at that point.
164. According to the IVT, cos x − sin x − x = 2 has a
solution over the interval [−1, 1].
vi. f is left continuous but not right continuous at
vii.
In the following exercises, suppose y = f (x) is defined for
x = a.
ii. f has an infinite discontinuity at x = −6.
iii. f (−6) = 3
iv.
and
x → 3+
x→0
4
156. Let f (x) = x 2 − 1 for x ≠ − 1, 1.
x −1
a. Sketch the graph of f.
b. Is it possible to find values k 1 and k 2 such that
f (−1) = k and f (1) = k 2, and that makes f (x)
157.
lim f (x) = 2.
lim f (x) = −∞
x → 3−
159. Discontinuous at x = 1 with
155. Let f (x) = ⎨ 3
.
⎩x , x < 1
which makes
hold:
f (x) = +∞
165. If f (x) is continuous such that f (a) and f (b) have
opposite signs, then f (x) = 0 has exactly one solution in
⎡
⎣
a, b⎤⎦.
166.
2
The function f (x) = x −2 4x + 3 is continuous
over the interval [0, 3].
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x −1
Chapter 2 | Limits
167.
f (x)
If
f (a), f (b) > 0,
⎡
⎣
193
is
continuous
everywhere
and
then there is no root of f (x) in the
⎤
⎦
interval a, b .
[T] The following problems consider the scalar form of
Coulomb’s law, which describes the electrostatic force
between two point charges, such as electrons. It is given by
the equation F(r) = k e
|q 1 q 2| ,
r
2
where k e is Coulomb’s
constant, q i are the magnitudes of the charges of the two
particles, and r is the distance between the two particles.
168. To simplify the calculation of a model with many
interacting particles, after some threshold value r = R,
we approximate F as zero.
a. Explain the physical reasoning behind this
assumption.
b. What is the force equation?
c. Evaluate the force F using both Coulomb’s law
and our approximation, assuming two protons with
a
charge
magnitude
of
1.6022 × 10 −19 coulombs (C),
and
the
Coulomb constant k e = 8.988 × 10 9 Nm 2/C 2 are
1 m apart. Also, assume R < 1 m. How much
inaccuracy does our approximation generate? Is our
approximation reasonable?
d. Is there any finite value of R for which this system
remains continuous at R?
169. Instead of making the force 0 at R, instead we let
the force be 10−20 for r ≥ R. Assume two protons, which
have a magnitude of charge 1.6022 × 10 −19 C, and the
Coulomb constant k e = 8.988 × 10 9 Nm 2/C 2. Is there a
value R that can make this system continuous? If so, find
it.
Recall the discussion on spacecraft from the chapter
opener. The following problems consider a rocket launch
from Earth’s surface. The force of gravity on the rocket is
given by F(d) = − mk/d 2, where m is the mass of the
rocket, d is the distance of the rocket from the center of
Earth, and k is a constant.
170. [T] Determine the value and units of k given that
the mass of the rocket on Earth is 3 million kg. (Hint: The
distance from the center of Earth to its surface is 6378 km.)
171. [T] After a certain distance D has passed, the
gravitational effect of Earth becomes quite negligible, so
we can approximate the force function by
⎧− mk if d < D
. Find the necessary condition
F(d) = ⎨ d 2
⎩10,000 if d ≥ D
D such that the force function remains continuous.
172. As the rocket travels away from Earth’s surface, there
is a distance D where the rocket sheds some of its mass,
since it no longer needs the excess fuel storage. We can
⎧ m1 k
⎪− 2 if d < D
d
write this function as F(d) = ⎨
. Is there
m
⎪− 2 k if d ≥ D
⎩ d2
a D value such that this function is continuous, assuming
m1 ≠ m2 ?
Prove the following functions are continuous everywhere
173.
f (θ) = sinθ
174. g(x) = |x|
⎧0
175. Where is f (x) = ⎨
⎩1
if x is irrational
continuous?
if x is rational
194
Chapter 2 | Limits
2.5 | The Precise Definition of a Limit
Learning Objectives
2.5.1
2.5.2
2.5.3
2.5.4
Describe the epsilon-delta definition of a limit.
Apply the epsilon-delta definition to find the limit of a function.
Describe the epsilon-delta definitions of one-sided limits and infinite limits.
Use the epsilon-delta definition to prove the limit laws.
By now you have progressed from the very informal definition of a limit in the introduction of this chapter to the intuitive
understanding of a limit. At this point, you should have a very strong intuitive sense of what the limit of a function means
and how you can find it. In this section, we convert this intuitive idea of a limit into a formal definition using precise
mathematical language. The formal definition of a limit is quite possibly one of the most challenging definitions you will
encounter early in your study of calculus; however, it is well worth any effort you make to reconcile it with your intuitive
notion of a limit. Understanding this definition is the key that opens the door to a better understanding of calculus.
Quantifying Closeness
Before stating the formal definition of a limit, we must introduce a few preliminary ideas. Recall that the distance between
two points a and b on a number line is given by |a − b|.
• The statement | f (x) − L| < ε may be interpreted as: The distance between f (x) and L is less than ε.
• The statement 0 < |x − a| < δ may be interpreted as: x ≠ a and the distance between x and a is less than δ.
It is also important to look at the following equivalences for absolute value:
• The statement | f (x) − L| < ε is equivalent to the statement L − ε < f (x) < L + ε.
• The statement 0 < |x − a| < δ is equivalent to the statement a − δ < x < a + δ and x ≠ a.
With these clarifications, we can state the formal epsilon-delta definition of the limit.
Definition
Let f (x) be defined for all x ≠ a over an open interval containing a. Let L be a real number. Then
lim f (x) = L
x→a
if, for every ε > 0, there exists a δ > 0, such that if 0 < |x − a| < δ, then | f (x) − L| < ε.
This definition may seem rather complex from a mathematical point of view, but it becomes easier to understand if we
break it down phrase by phrase. The statement itself involves something called a universal quantifier (for every ε > 0), an
existential quantifier (there exists a δ > 0), and, last, a conditional statement (if 0 < |x − a| < δ, then | f (x) − L| < ε).
Let’s take a look at Table 2.9, which breaks down the definition and translates each part.
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Chapter 2 | Limits
195
Definition
Translation
1. For every ε > 0,
1. For every positive distance ε from L,
2. there exists a δ > 0,
2. There is a positive distance δ from a,
3. such that
3. such that
4. if 0 < |x − a| < δ, then
4. if x is closer than δ to a and x ≠ a, then f (x) is closer than
| f (x) − L| < ε.
ε to L.
Table 2.9 Translation of the Epsilon-Delta Definition of the Limit
We can get a better handle on this definition by looking at the definition geometrically. Figure 2.39 shows possible values
of δ for various choices of ε > 0 for a given function f (x), a number a, and a limit L at a. Notice that as we choose
smaller values of ε (the distance between the function and the limit), we can always find a δ small enough so that if we
have chosen an x value within δ of a, then the value of f (x) is within ε of the limit L.
Figure 2.39 These graphs show possible values of δ , given successively smaller choices of ε.
Visit the following applet to experiment with finding values of δ for selected values of ε:
• The epsilon-delta definition of limit (http://www.openstaxcollege.org/l/20_epsilondelt)
Example 2.39 shows how you can use this definition to prove a statement about the limit of a specific function at a
specified value.
Example 2.39
Proving a Statement about the Limit of a Specific Function
Prove that lim (2x + 1) = 3.
x→1
Solution
196
Chapter 2 | Limits
Let ε > 0.
The first part of the definition begins “For every ε > 0.” This means we must prove that whatever follows is true
no matter what positive value of ε is chosen. By stating “Let ε > 0,” we signal our intent to do so.
Choose δ = ε .
2
The definition continues with “there exists a δ > 0. ” The phrase “there exists” in a mathematical statement is
always a signal for a scavenger hunt. In other words, we must go and find δ. So, where exactly did δ = ε/2
come from? There are two basic approaches to tracking down δ. One method is purely algebraic and the other is
geometric.
We begin by tackling the problem from an algebraic point of view. Since ultimately we want |(2x + 1) − 3| < ε,
we begin by manipulating this expression: |(2x + 1) − 3| < ε is equivalent to |2x − 2| < ε, which in turn
is equivalent to |2||x − 1| < ε. Last, this is equivalent to |x − 1| < ε/2. Thus, it would seem that δ = ε/2 is
appropriate.
We may also find δ through geometric methods. Figure 2.40 demonstrates how this is done.
Figure 2.40 This graph shows how we find δ geometrically.
Assume 0 < |x − 1| < δ. When δ has been chosen, our goal is to show that if 0 < |x − 1| < δ,
|(2x + 1) − 3| < ε.
then
To prove any statement of the form “If this, then that,” we begin by assuming “this” and
trying to get “that.”
Thus,
|(2x + 1) − 3| = |2x − 2|
= |2(x − 1)|
= |2||x − 1|
= 2|x − 1|
< 2·δ
= 2· ε = ε
2
property of absolute value
|2| = 2
here’s where we use the assumption that 0 < |x − 1| < δ
here’s where we use our choice of δ = ε/2
Analysis
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Chapter 2 | Limits
197
In this part of the proof, we started with |(2x + 1) − 3| and used our assumption 0 < |x − 1| < δ in a key part
of the chain of inequalities to get |(2x + 1) − 3| to be less than ε. We could just as easily have manipulated the
assumed inequality 0 < |x − 1| < δ to arrive at |(2x + 1) − 3| < ε as follows:
0 < | x − 1| < δ ⇒ | x − 1| < δ
⇒ −δ< x−1<δ
⇒ − ε < x−1< ε
2
2
⇒ − ε < 2x − 2 < ε
⇒ − ε < 2x − 2 < ε
⇒ |2x − 2| < ε
⇒ |(2x + 1) − 3| < ε.
Therefore, lim (2x + 1) = 3. (Having completed the proof, we state what we have accomplished.)
x→1
After removing all the remarks, here is a final version of the proof:
Let ε > 0.
Choose δ = ε/2.
Assume 0 < |x − 1| < δ.
Thus,
|(2x + 1) − 3| = |2x − 2|
= |2(x − 1)|
= |2||x − 1|
= 2|x − 1|
< 2·δ
= 2· ε
2
= ε.
Therefore, lim (2x + 1) = 3.
x→1
The following Problem-Solving Strategy summarizes the type of proof we worked out in Example 2.39.
Problem-Solving Strategy: Proving That xlim
f(x) = L for a Specific Function f(x)
→a
1. Let’s begin the proof with the following statement: Let ε > 0.
2. Next, we need to obtain a value for δ. After we have obtained this value, we make the following statement,
filling in the blank with our choice of δ : Choose δ = _______.
3. The next statement in the proof should be (at this point, we fill in our given value for a):
Assume 0 < |x − a| < δ.
4. Next, based on this assumption, we need to show that | f (x) − L| < ε, where f (x) and L are our function
f (x) and our limit L. At some point, we need to use 0 < |x − a| < δ.
5. We conclude our proof with the statement: Therefore, xlim
f (x) = L.
→a
198
Chapter 2 | Limits
Example 2.40
Proving a Statement about a Limit
Complete the proof that
lim (4x + 1) = −3 by filling in the blanks.
x → −1
Let _____.
Choose δ = _______.
Assume 0 < |x − _______| < δ.
Thus, |________ − ________| = _____________________________________ε.
Solution
We begin by filling in the blanks where the choices are specified by the definition. Thus, we have
Let ε > 0.
Choose δ = _______.
Assume 0 < |x − (−1)| < δ. (or equivalently, 0 < |x + 1| < δ.)
Thus, |(4x + 1) − (−3)| = |4x + 4| = |4||x + 1| < 4δ_______ε.
Focusing on the final line of the proof, we see that we should choose δ = ε .
4
We now complete the final write-up of the proof:
Let ε > 0.
Choose δ = ε .
4
Assume 0 < |x − (−1)| < δ (or equivalently, 0 < |x + 1| < δ.)
Thus, |(4x + 1) − (−3)| = |4x + 4| = |4||x + 1| < 4δ = 4(ε/4) = ε.
2.27
Complete the proof that lim (3x − 2) = 4 by filling in the blanks.
x→2
Let _______.
Choose δ = _______.
Assume 0 < |x − ____| < ____.
Thus,
|_______ − ____| = ______________________________ε.
Therefore, lim (3x − 2) = 4.
x→2
In Example 2.39 and Example 2.40, the proofs were fairly straightforward, since the functions with which we were
working were linear. In Example 2.41, we see how to modify the proof to accommodate a nonlinear function.
Example 2.41
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Chapter 2 | Limits
199
Proving a Statement about the Limit of a Specific Function (Geometric Approach)
Prove that lim x 2 = 4.
x→2
Solution
1. Let ε > 0. The first part of the definition begins “For every ε > 0,” so we must prove that whatever
follows is true no matter what positive value of ε is chosen. By stating “Let ε > 0,” we signal our intent
to do so.
2. Without loss of generality, assume ε ≤ 4. Two questions present themselves: Why do we want ε ≤ 4
and why is it okay to make this assumption? In answer to the first question: Later on, in the process of
solving for δ, we will discover that δ involves the quantity 4 − ε. Consequently, we need ε ≤ 4. In
answer to the second question: If we can find δ > 0 that “works” for ε ≤ 4, then it will “work” for any
ε > 4 as well. Keep in mind that, although it is always okay to put an upper bound on ε, it is never okay
to put a lower bound (other than zero) on ε.
⎧
⎫
3. Choose δ = min⎨⎩2 − 4 − ε, 4 + ε − 2⎬⎭. Figure 2.41 shows how we made this choice of δ.
Figure 2.41 This graph shows how we find δ geometrically for a given ε
for the proof in Example 2.41.
4. We must show: If 0 < |x − 2| < δ, then |x 2 − 4| < ε, so we must begin by assuming
0 < |x − 2| < δ.
We
don’t
really
need
0 < |x − 2|
(in
other
words,
x ≠ 2)
for
this
δ≥2− 4−ε
and
proof.
Since
0 < |x − 2| < δ ⇒ |x − 2| < δ, it is okay to drop 0 < |x − 2|.
|x − 2| < δ.
Hence,
−δ < x − 2 < δ.
Recall
that
⎧
⎨
⎩
⎫
⎬
⎭
δ = min 2 − 4 − ε, 4 + ε − 2 .
Thus,
consequently
− ⎛⎝2 − 4 − ε⎞⎠ ≤ − δ. We also use δ ≤ 4 + ε − 2 here. We might ask at this point: Why did we
substitute 2 − 4 − ε for δ on the left-hand side of the inequality and
4 + ε − 2 on the right-hand
side of the inequality? If we look at Figure 2.41, we see that 2 − 4 − ε corresponds to the distance on
200
Chapter 2 | Limits
the left of 2 on the x-axis and 4 + ε − 2 corresponds to the distance on the right. Thus,
− ⎛⎝2 − 4 − ε⎞⎠ ≤ − δ < x − 2 < δ ≤ 4 + ε − 2.
We simplify the expression on the left:
−2 + 4 − ε < x − 2 < 4 + ε − 2.
Then, we add 2 to all parts of the inequality:
4 − ε < x < 4 + ε.
We square all parts of the inequality. It is okay to do so, since all parts of the inequality are positive:
4 − ε < x 2 < 4 + ε.
We subtract 4 from all parts of the inequality:
−ε < x 2 − 4 < ε.
Last,
|x 2 − 4| < ε.
5. Therefore,
lim x 2 = 4.
x→2
2.28
Find δ corresponding to ε > 0 for a proof that lim x = 3.
x→9
The geometric approach to proving that the limit of a function takes on a specific value works quite well for some functions.
Also, the insight into the formal definition of the limit that this method provides is invaluable. However, we may also
approach limit proofs from a purely algebraic point of view. In many cases, an algebraic approach may not only provide
us with additional insight into the definition, it may prove to be simpler as well. Furthermore, an algebraic approach is the
primary tool used in proofs of statements about limits. For Example 2.42, we take on a purely algebraic approach.
Example 2.42
Proving a Statement about the Limit of a Specific Function (Algebraic Approach)
Prove that
lim ⎛⎝x 2 − 2x + 3⎞⎠ = 6.
x → −1
Solution
Let’s use our outline from the Problem-Solving Strategy:
1. Let ε > 0.
2. Choose δ = min{1, ε/5}. This choice of δ may appear odd at first glance, but it was obtained by
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Chapter 2 | Limits
201
taking a look at our ultimate desired inequality:
|x
⎛ 2
⎝
|
− 2x + 3⎞⎠ − 6 < ε. This inequality is equivalent
ε
is very strong.
x−3
Unfortunately, our choice of δ must depend on ε only and no other variable. If we can replace |x − 3| by
a numerical value, our problem can be resolved. This is the place where assuming δ ≤ 1 comes into play.
The choice of δ ≤ 1 here is arbitrary. We could have just as easily used any other positive number. In
some proofs, greater care in this choice may be necessary. Now, since δ ≤ 1 and |x + 1| < δ ≤ 1, we
to |x + 1| · |x − 3| < ε. At this point, the temptation simply to choose δ =
are able to show that |x − 3| < 5. Consequently, |x + 1| · |x − 3| < |x + 1| · 5. At this point we realize
that we also need δ ≤ ε/5. Thus, we choose δ = min{1, ε/5}.
3. Assume 0 < |x + 1| < δ. Thus,
|x + 1| < 1 and |x + 1| < 5ε .
Since |x + 1| < 1, we may conclude that −1 < x + 1 < 1. Thus, by subtracting 4 from all parts of the
inequality, we obtain −5 < x − 3 < −1. Consequently, |x − 3| < 5. This gives us
|x
⎛ 2
⎝
|
− 2x + 3⎞⎠ − 6 = |x + 1| · |x − 3| < ε · 5 = ε.
5
Therefore,
lim ⎛⎝x 2 − 2x + 3⎞⎠ = 6.
x → −1
2.29
Complete the proof that lim x 2 = 1.
x→1
Let ε > 0; choose δ = min{1, ε/3}; assume 0 < |x − 1| < δ.
Since |x − 1| < 1, we may conclude that −1 < x − 1 < 1. Thus, 1 < x + 1 < 3. Hence, |x + 1| < 3.
You will find that, in general, the more complex a function, the more likely it is that the algebraic approach is the easiest to
apply. The algebraic approach is also more useful in proving statements about limits.
Proving Limit Laws
We now demonstrate how to use the epsilon-delta definition of a limit to construct a rigorous proof of one of the limit laws.
The triangle inequality is used at a key point of the proof, so we first review this key property of absolute value.
Definition
The triangle inequality states that if a and b are any real numbers, then |a + b| ≤ |a| + |b|.
Proof
⎛
We prove the following limit law: If xlim
f (x) = L and xlim
g(x) = M, then xlim
f (x) + g(x)⎞⎠ = L + M.
→a
→a
→ a⎝
Let ε > 0.
Choose δ 1 > 0 so that if 0 < |x − a| < δ 1, then | f (x) − L| < ε/2.
Choose δ 2 > 0 so that if 0 < |x − a| < δ 2, then |g(x) − M | < ε/2.
202
Chapter 2 | Limits
⎧
⎨
⎩
⎫
⎬
⎭
Choose δ = min δ 1, δ 2 .
Assume 0 < |x − a| < δ.
Thus,
0 < |x − a| < δ 1 and 0 < |x − a| < δ 2.
Hence,
| f (x) + g(x)
⎛
⎝
⎞
⎠
− (L + M)| = |⎛⎝ f (x) − L⎞⎠ + ⎛⎝g(x) − M ⎞⎠|
≤ | f (x) − L| + |g(x) − M |
< ε + ε = ε.
2 2
□
We now explore what it means for a limit not to exist. The limit xlim
f (x) does not exist if there is no real number L for
→a
which xlim
f (x) = L. Thus, for all real numbers L, xlim
f (x) ≠ L. To understand what this means, we look at each part
→a
→a
of the definition of xlim
f (x) = L together with its opposite. A translation of the definition is given in Table 2.10.
→a
Definition
Opposite
1. For every ε > 0,
1. There exists ε > 0 so that
2. there exists a δ > 0, so that
2. for every δ > 0,
3. if 0 < |x − a| < δ, then
3. There is an x satisfying 0 < |x − a| < δ so that
| f (x) − L| < ε.
| f (x) − L| ≥ ε.
Table 2.10 Translation of the Definition of xlim
f (x) = L and its Opposite
→a
Finally, we may state what it means for a limit not to exist. The limit xlim
f (x) does not exist if for every real number L,
→a
there exists a real number ε > 0 so that for all δ > 0, there is an x satisfying 0 < |x − a| < δ, so that | f (x) − L| ≥ ε.
Let’s apply this in Example 2.43 to show that a limit does not exist.
Example 2.43
Showing That a Limit Does Not Exist
Show that lim |xx| does not exist. The graph of f (x) = |x|/x is shown here:
x→0
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Chapter 2 | Limits
203
Solution
Suppose that L is a candidate for a limit. Choose ε = 1/2.
Let δ > 0. Either L ≥ 0 or L < 0. If L ≥ 0, then let x = − δ/2. Thus,
|
|
|x − 0| = − δ − 0 = δ < δ
2
2
and
| |
|− | − L = |−1 − L| = L + 1 ≥ 1 > 12 = ε.
δ
2
− 2δ
On the other hand, if L < 0, then let x = δ/2. Thus,
| |
|x − 0| = δ − 0 = δ < δ
2
2
and
| |
| | − L = |1 − L| = |L| + 1 ≥ 1 > 12 = ε.
δ
2
δ
2
Thus, for any value of L, lim |xx| ≠ L.
x→0
One-Sided and Infinite Limits
Just as we first gained an intuitive understanding of limits and then moved on to a more rigorous definition of a limit,
we now revisit one-sided limits. To do this, we modify the epsilon-delta definition of a limit to give formal epsilon-delta
definitions for limits from the right and left at a point. These definitions only require slight modifications from the definition
of the limit. In the definition of the limit from the right, the inequality 0 < x − a < δ replaces 0 < |x − a| < δ, which
ensures that we only consider values of x that are greater than (to the right of) a. Similarly, in the definition of the limit from
the left, the inequality −δ < x − a < 0 replaces 0 < |x − a| < δ, which ensures that we only consider values of x that
are less than (to the left of) a.
Definition
Limit from the Right: Let f (x) be defined over an open interval of the form (a, b) where a < b. Then,
204
Chapter 2 | Limits
lim f (x) = L
x → a+
if for every ε > 0, there exists a δ > 0 such that if 0 < x − a < δ, then | f (x) − L| < ε.
Limit from the Left: Let f (x) be defined over an open interval of the form (b, c) where b < c. Then,
lim
x → a−
f (x) = L
if for every ε > 0, there exists a δ > 0 such that if −δ < x − a < 0, then | f (x) − L| < ε.
Example 2.44
Proving a Statement about a Limit From the Right
Prove that
lim
x → 4+
x − 4 = 0.
Solution
Let ε > 0.
Choose δ = ε 2. Since we ultimately want
2
|
|
x − 4 − 0 < ε, we manipulate this inequality to get x − 4 < ε
or, equivalently, 0 < x − 4 < ε , making δ = ε
2
a clear choice. We may also determine δ geometrically, as
shown in Figure 2.42.
Figure 2.42 This graph shows how we find δ for the proof in
Example 2.44.
Assume 0 < x − 4 < δ. Thus, 0 < x − 4 < ε 2. Hence, 0 < x − 4 < ε. Finally,
Therefore,
2.30
lim
x → 4+
|
|
x − 4 − 0 < ε.
x − 4 = 0.
Find δ corresponding to ε for a proof that
lim
x → 1−
1 − x = 0.
We conclude the process of converting our intuitive ideas of various types of limits to rigorous formal definitions by
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Chapter 2 | Limits
205
pursuing a formal definition of infinite limits. To have xlim
f (x) = +∞, we want the values of the function f (x)
→a
to get larger and larger as x approaches a. Instead of the requirement that | f (x) − L| < ε for arbitrarily small ε when
0 < |x − a| < δ for small enough δ, we want f (x) > M for arbitrarily large positive M when 0 < |x − a| < δ for small
enough δ. Figure 2.43 illustrates this idea by showing the value of δ for successively larger values of M.
Figure 2.43 These graphs plot values of δ for M to show that lim f (x) = +∞.
x→a
Definition
Let f (x) be defined for all x ≠ a in an open interval containing a. Then, we have an infinite limit
lim f (x) = +∞
x→a
if for every M > 0, there exists δ > 0 such that if 0 < |x − a| < δ, then f (x) > M.
Let f (x) be defined for all x ≠ a in an open interval containing a. Then, we have a negative infinite limit
lim f (x) = −∞
x→a
if for every M > 0, there exists δ > 0 such that if 0 < |x − a| < δ, then f (x) < −M.
206
Chapter 2 | Limits
2.5 EXERCISES
In the following exercises, write the appropriate ε − δ
definition for each of the given statements.
176. xlim
f (x) = N
→a
177.
lim g(t) = M
t→b
178. xlim
h(x) = L
→c
179. xlim
φ(x) = A
→a
The
following
graph
of
the
function
f
satisfies
lim f (x) = 2. In the following exercises, determine a
x→2
value of δ > 0 that satisfies each statement.
182. If 0 < |x − 3| < δ, then | f (x) + 1| < 1.
183. If 0 < |x − 3| < δ, then | f (x) + 1| < 2.
The
following
graph
of
the
function
f
satisfies
lim f (x) = 2. In the following exercises, for each value
x→3
of ε, find a value of δ > 0 such that the precise definition
of limit holds true.
180. If 0 < |x − 2| < δ, then | f (x) − 2| < 1.
181. If 0 < |x − 2| < δ, then | f (x) − 2| < 0.5.
The
following
graph
of
the
function
f
satisfies
lim f (x) = −1. In the following exercises, determine a
x→3
value of δ > 0 that satisfies each statement.
184. ε = 1.5
185. ε = 3
[T] In the following exercises, use a graphing calculator to
find a number δ such that the statements hold true.
186.
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|sin(2x) − 12| < 0.1,
|
|
whenever x − π < δ
12
Chapter 2 | Limits
187.
|
207
|
x − 4 − 2 < 0.1, whenever |x − 8| < δ
In the following exercises, use the precise definition of
limit to prove the given limits.
188.
191.
192.
x→0
2
lim 2x − 3x − 2 = 5
x−2
x→2
203.
lim x 4 = 0
into two cases, x rational and x irrational.)
lim (x 2 + 2x) = 8
204. Using the function from the previous exercise, use
the precise definition of limits to show that xlim
f (x) does
→a
x→0
x→2
lim
x → 5−
5−x=0
not exist for a ≠ 0.
For the following exercises, suppose that xlim
f (x) = L
→a
and xlim
g(x) = M both exist. Use the precise definition
→a
of limits to prove the following limit laws:
lim f (x) = 3, where f (x) =
x → 1−
⎧5x
⎨
⎩7x
− 2, if x < 1
.
− 1, if x ≥ 1
In the following exercises, use the precise definition of
limit to prove the given infinite limits.
198.
Using precise definitions of limits, determine
⎧x if x is rational
lim f (x) for f (x) = ⎨⎩
. (Hint: Break
0 if x is irrational
⎛
205. xlim
f (x) − g(x)⎞⎠ = L − M
→ a⎝
206.
lim 1 = ∞
x → 0 x2
lim
3
x → −1 (x + 1) 2
lim −
x→2
=∞
1
= −∞
(x − 2) 2
199. An engineer is using a machine to cut a flat square
of Aerogel of area 144 cm2. If there is a maximum error
tolerance in the area of 8 cm2, how accurately must the
engineer cut on the side, assuming all sides have the same
length? How do these numbers relate to δ, ε, a, and L?
200. Use the precise definition of limit to prove that the
following limit does not exist: lim |x − 1| .
x→1
x−1
lim ⎡⎣c f (x)⎤⎦ = cL for any real constant c (Hint:
x→a
Consider two cases: c = 0 and c ≠ 0.)
207.
197.
⎧1 if x is rational
f (x) = ⎨⎩
.
0 if x is irrational
x→0
⎧8x − 3, if x < 0
.
lim f (x) = −2, where f (x) = ⎨
+
⎩4x − 2, if x ≥ 0
x→0
196.
Using precise definitions of limits, prove that
(Hint: Think about how you can always choose a rational
number 0 < r < d, but | f (r) − 0| = 1.)
194.
195.
x→0
function. (Hint: Try any δ < 1.)
lim f (x) does not exist:
In the following exercises, use the precise definition of
limit to prove the given one-sided limits.
193.
Using precise definitions of limits, prove that
lim f (x) does not exist, given that f (x) is the ceiling
202.
lim (5x + 8) = 18
x→2
2
189. lim x − 9 = 6
x→3 x−3
190.
201.
lim ⎡⎣ f (x)g(x)⎤⎦ = LM. (Hint: | f (x)g(x) − LM | =
x→a
| f (x)g(x) − f (x)M + f (x)M − LM| ≤ | f (x)||g(x) − M| + |M|| f (x) − L|.)
208
Chapter 2 | Limits
CHAPTER 2 REVIEW
KEY TERMS
average velocity the change in an object’s position divided by the length of a time period; the average velocity of an
object over a time interval [t, a] (if t < a or [a, t] if t > a) , with a position given by s(t), that is
v ave =
s(t) − s(a)
t−a
c f (x) = c · xlim
f (x) = cL
constant multiple law for limits the limit law xlim
→a
→a
continuity at a point A function f (x) is continuous at a point a if and only if the following three conditions are
satisfied: (1) f (a) is defined, (2) xlim
f (x) exists, and (3) xlim
f (x) = f (a)
→a
→a
continuity from the left A function is continuous from the left at b if
lim f (x) = f (b)
x → b−
continuity from the right A function is continuous from the right at a if
lim f (x) = f (a)
x → a+
continuity over an interval a function that can be traced with a pencil without lifting the pencil; a function is
continuous over an open interval if it is continuous at every point in the interval; a function f (x) is continuous over a
closed interval of the form ⎡⎣a, b⎤⎦ if it is continuous at every point in (a, b), and it is continuous from the right at a
and from the left at b
difference law for limits the limit law lim ⎛⎝ f (x) − g(x)⎞⎠ = lim f (x) − lim g(x) = L − M
x→a
x→a
x→a
differential calculus the field of calculus concerned with the study of derivatives and their applications
discontinuity at a point A function is discontinuous at a point or has a discontinuity at a point if it is not continuous at
the point
f (x) = L if for every ε > 0,
epsilon-delta definition of the limit xlim
→a
0 < |x − a| < δ, then | f (x) − L| < ε
infinite discontinuity An infinite discontinuity occurs at a point a if
there exists a δ > 0 such that if
lim f (x) = ±∞ or
x → a−
lim f (x) = ±∞
x → a+
infinite limit A function has an infinite limit at a point a if it either increases or decreases without bound as it approaches
a
instantaneous velocity The instantaneous velocity of an object with a position function that is given by s(t) is the
value that the average velocities on intervals of the form [t, a] and [a, t] approach as the values of t move closer to
a, provided such a value exists
integral calculus the study of integrals and their applications
Intermediate Value Theorem Let f be continuous over a closed bounded interval ⎡⎣a, b⎤⎦; if z is any real number
between f (a) and f (b), then there is a number c in ⎡⎣a, b⎤⎦ satisfying f (c) = z
intuitive definition of the limit If all values of the function f (x) approach the real number L as the values of x( ≠ a)
approach a, f (x) approaches L
jump discontinuity A jump discontinuity occurs at a point a if
lim
x → a−
f (x) ≠ lim f (x)
lim
x → a−
f (x) and
lim f (x) both exist, but
x → a+
x → a+
limit the process of letting x or t approach a in an expression; the limit of a function f (x) as x approaches a is the value
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Chapter 2 | Limits
209
that f (x) approaches as x approaches a
limit laws the individual properties of limits; for each of the individual laws, let f (x) and g(x) be defined for all x ≠ a
over some open interval containing a; assume that L and M are real numbers so that xlim
f (x) = L and
→a
lim g(x) = M; let c be a constant
x→a
multivariable calculus the study of the calculus of functions of two or more variables
one-sided limit A one-sided limit of a function is a limit taken from either the left or the right
power law for limits
⎛
⎞
n
⎛
f (x)⎠ = L n for every positive integer n
the limit law xlim
f (x)⎞⎠ n = ⎝xlim
→a
→ a⎝
product law for limits the limit law lim ⎛⎝ f (x) · g(x)⎞⎠ = lim f (x) · lim g(x) = L · M
x→a
x→a
x→a
quotient law for limits
the limit law xlim
→a
f (x)
f (x) xlim
= →a
= L for M ≠ 0
M
g(x) xlim
g(x)
→a
f (x)
removable discontinuity A removable discontinuity occurs at a point a if f (x) is discontinuous at a, but xlim
→a
exists
root law for limits the limit law lim n f (x) = n lim f (x) = n L for all L if n is odd and for L ≥ 0 if n is even
x→a
x→a
secant A secant line to a function f (x) at a is a line through the point ⎛⎝a, f (a)⎞⎠ and another point on the function; the
slope of the secant line is given by m sec =
squeeze theorem states that if
f (x) − f (a)
x−a
f (x) ≤ g(x) ≤ h(x) for all x ≠ a over an open interval containing a and
lim f (x) = L = xlim
h(x) where L is a real number, then xlim
g(x) = L
x→a
→a
→a
sum law for limits The limit law lim ⎛⎝ f (x) + g(x)⎞⎠ = lim f (x) + lim g(x) = L + M
x→a
x→a
x→a
tangent A tangent line to the graph of a function at a point ⎛⎝a, f (a)⎞⎠ is the line that secant lines through ⎛⎝a, f (a)⎞⎠
approach as they are taken through points on the function with x-values that approach a; the slope of the tangent line
to a graph at a measures the rate of change of the function at a
triangle inequality If a and b are any real numbers, then |a + b| ≤ |a| + |b|
vertical asymptote A function has a vertical asymptote at x = a if the limit as x approaches a from the right or left is
infinite
KEY EQUATIONS
• Slope of a Secant Line
m sec =
f (x) − f (a)
x−a
• Average Velocity over Interval [a, t]
v ave =
s(t) − s(a)
t−a
• Intuitive Definition of the Limit
lim f (x) = L
x→a
• Two Important Limits
lim x = a
x→a
lim c = c
x→a
• One-Sided Limits
210
Chapter 2 | Limits
lim
x → a−
f (x) = L
lim f (x) = L
x → a+
• Infinite Limits from the Left
lim f (x) = +∞
lim f (x) = −∞
x → a−
x → a−
• Infinite Limits from the Right
lim f (x) = −∞
lim f (x) = +∞
x → a+
x → a+
• Two-Sided Infinite Limits
lim f (x) = +∞ : lim− f (x) = +∞ and
x→a
x→a
lim f (x) = −∞ : lim− f (x) = −∞ and
x→a
x→a
lim f (x) = +∞
x → a+
lim f (x) = −∞
x → a+
• Basic Limit Results
lim x = a xlim
c=c
→a
x→a
• Important Limits
lim sinθ = 0
θ→0
lim cosθ = 1
θ→0
lim sinθ = 1
θ
1
lim − cosθ = 0
θ
θ→0
θ→0
KEY CONCEPTS
2.1 A Preview of Calculus
• Differential calculus arose from trying to solve the problem of determining the slope of a line tangent to a curve at a
point. The slope of the tangent line indicates the rate of change of the function, also called the derivative. Calculating
a derivative requires finding a limit.
• Integral calculus arose from trying to solve the problem of finding the area of a region between the graph of a
function and the x-axis. We can approximate the area by dividing it into thin rectangles and summing the areas of
these rectangles. This summation leads to the value of a function called the integral. The integral is also calculated
by finding a limit and, in fact, is related to the derivative of a function.
• Multivariable calculus enables us to solve problems in three-dimensional space, including determining motion in
space and finding volumes of solids.
2.2 The Limit of a Function
• A table of values or graph may be used to estimate a limit.
• If the limit of a function at a point does not exist, it is still possible that the limits from the left and right at that point
may exist.
• If the limits of a function from the left and right exist and are equal, then the limit of the function is that common
value.
• We may use limits to describe infinite behavior of a function at a point.
2.3 The Limit Laws
• The limit laws allow us to evaluate limits of functions without having to go through step-by-step processes each
time.
• For polynomials and rational functions, xlim
f (x) = f (a).
→a
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Chapter 2 | Limits
211
• You can evaluate the limit of a function by factoring and canceling, by multiplying by a conjugate, or by simplifying
a complex fraction.
• The squeeze theorem allows you to find the limit of a function if the function is always greater than one function
and less than another function with limits that are known.
2.4 Continuity
• For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the
value of the function at that point must equal the value of the limit at that point.
• Discontinuities may be classified as removable, jump, or infinite.
• A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over
a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
• The composite function theorem states: If
f (x)
is continuous at L and
g(x)⎞⎠ = f (L).
lim f ⎛⎝g(x)⎞⎠ = f ⎛⎝xlim
→a
lim g(x) = L,
x→a
then
x→a
• The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function
takes on every value between the values at its endpoints.
2.5 The Precise Definition of a Limit
• The intuitive notion of a limit may be converted into a rigorous mathematical definition known as the epsilon-delta
definition of the limit.
• The epsilon-delta definition may be used to prove statements about limits.
• The epsilon-delta definition of a limit may be modified to define one-sided limits.
CHAPTER 2 REVIEW EXERCISES
True or False. In the following exercises, justify your
answer with a proof or a counterexample.
208. A function has to be continuous at x = a if the
lim f (x) exists.
x→a
212. Using the graph, find each limit or explain why the
limit does not exist.
a.
lim f (x)
x → −1
b.
c.
x.
209. You can use the quotient rule to evaluate lim sin
x→0 x
d.
lim f (x)
x→1
lim f (x)
x → 0+
lim f (x)
x→2
210. If there is a vertical asymptote at x = a for the
function f (x), then f is undefined at the point x = a.
211. If xlim
f (x) does not exist, then f is undefined at the
→a
point x = a.
In the following exercises, evaluate the limit algebraically
or explain why the limit does not exist.
213.
2
lim 2x − 3x − 2
x−2
x→2
212
Chapter 2 | Limits
lim (8x + 16) = 24
lim 3x 2 − 2x + 4
228.
215.
3
2
lim x − 2x − 1
3x
−
2
x→3
229.
216.
lim cot x
x → π/2 cos x
230. A ball is thrown into the air and the vertical position
is given by x(t) = −4.9t 2 + 25t + 5. Use the Intermediate
214.
217.
x→0
2
lim x + 25
x+5
x → −5
218.
2
2x − 8
lim 3x −
2
x→2
x −4
219.
2
lim x 3 − 1
x → 1x − 1
220.
221.
222.
2
lim x − 1
x−1
x→1
lim 4 − x
x−2
x→4
lim
x→4
1
x−2
In the following exercises, use the squeeze theorem to
prove the limit.
223.
224.
lim x 2 cos(2πx) = 0
x→0
lim x 3 sin ⎛⎝πx ⎞⎠ = 0
x→0
225.
Determine the domain such that the function
f (x) = x − 2 + xe x is continuous over its domain.
In the following exercises, determine the value of c such
that the function remains continuous. Draw your resulting
function to ensure it is continuous.
226.
⎧x 2 + 1, x > c
f (x) = ⎨
⎩2x, x ≤ c
227.
⎧ x + 1, x > −1
f (x) = ⎨ 2
⎩x + c, x ≤ − 1
In the following exercises, use the precise definition of
limit to prove the limit.
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x→1
lim x 3 = 0
x→0
Value Theorem to show that the ball must land on the
ground sometime between 5 sec and 6 sec after the throw.
231. A particle moving along a line has a displacement
according to the function x(t) = t 2 − 2t + 4, where x is
measured in meters and t is measured in seconds. Find the
average velocity over the time period t = [0, 2].
232.
From the previous exercises, estimate the
instantaneous velocity at t = 2 by checking the average
velocity within t = 0.01 sec.
Chapter 3 | Derivatives
213
3 | DERIVATIVES
Figure 3.1 The Hennessey Venom GT can go from 0 to 200 mph in 14.51 seconds. (credit: modification of work by Codex41,
Flickr)
Chapter Outline
3.1 Defining the Derivative
3.2 The Derivative as a Function
3.3 Differentiation Rules
3.4 Derivatives as Rates of Change
3.5 Derivatives of Trigonometric Functions
3.6 The Chain Rule
3.7 Derivatives of Inverse Functions
3.8 Implicit Differentiation
3.9 Derivatives of Exponential and Logarithmic Functions
Introduction
The Hennessey Venom GT is one of the fastest cars in the world. In 2014, it reached a record-setting speed of 270.49 mph.
It can go from 0 to 200 mph in 14.51 seconds. The techniques in this chapter can be used to calculate the acceleration the
Venom achieves in this feat (see Example 3.8.)
Calculating velocity and changes in velocity are important uses of calculus, but it is far more widespread than that. Calculus
is important in all branches of mathematics, science, and engineering, and it is critical to analysis in business and health as
214
Chapter 3 | Derivatives
well. In this chapter, we explore one of the main tools of calculus, the derivative, and show convenient ways to calculate
derivatives. We apply these rules to a variety of functions in this chapter so that we can then explore applications of these
techniques.
3.1 | Defining the Derivative
Learning Objectives
3.1.1
3.1.2
3.1.3
3.1.4
3.1.5
3.1.6
3.1.7
Recognize the meaning of the tangent to a curve at a point.
Calculate the slope of a tangent line.
Identify the derivative as the limit of a difference quotient.
Calculate the derivative of a given function at a point.
Describe the velocity as a rate of change.
Explain the difference between average velocity and instantaneous velocity.
Estimate the derivative from a table of values.
Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established
the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals.
Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton
(1643–1727) and the German Gottfried Leibniz (1646–1716), whose images appear in Figure 3.2. When we credit
Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first
to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of
predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to
have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it
seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation
that we commonly use today.
Figure 3.2 Newton and Leibniz are credited with developing calculus independently.
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Chapter 3 | Derivatives
215
Tangent Lines
We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of
a secant line to a function at a point (a, f (a)) to estimate the rate of change, or the rate at which one variable changes in
relation to another variable. We can obtain the slope of the secant by choosing a value of x near a and drawing a line
through the points (a, f (a)) and ⎛⎝x, f (x)⎞⎠, as shown in Figure 3.3. The slope of this line is given by an equation in the
form of a difference quotient:
m sec =
f (x) − f (a)
x−a .
We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing x with
a + h, where h is a value close to a. We can then calculate the slope of the line through the points (a, f (a)) and
(a + h, f (a + h)). In this case, we find the secant line has a slope given by the following difference quotient with
increment h:
m sec =
f (a + h) − f (a)
f (a + h) − f (a)
=
.
a+h−a
h
Definition
Let f be a function defined on an interval I containing a. If x ≠ a is in I, then
f (x) − f (a)
x−a
(3.1)
f (a + h) − f (a)
h
(3.2)
Q=
is a difference quotient.
Also, if h ≠ 0 is chosen so that a + h is in I, then
Q=
is a difference quotient with increment h.
View several Java applets (http://www.openstaxcollege.org/l/20_calcapplets) on the development of the
derivative.
These two expressions for calculating the slope of a secant line are illustrated in Figure 3.3. We will see that each of these
two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The
primary consideration in our choice usually depends on ease of calculation.
216
Chapter 3 | Derivatives
Figure 3.3 We can calculate the slope of a secant line in either of two ways.
In Figure 3.4(a) we see that, as the values of x approach a, the slopes of the secant lines provide better estimates of the
rate of change of the function at a. Furthermore, the secant lines themselves approach the tangent line to the function at
a, which represents the limit of the secant lines. Similarly, Figure 3.4(b) shows that as the values of h get closer to 0,
the secant lines also approach the tangent line. The slope of the tangent line at a is the rate of change of the function at a,
as shown in Figure 3.4(c).
Figure 3.4 The secant lines approach the tangent line (shown in green) as the second point approaches the first.
You can use this site (http://www.openstaxcollege.org/l/20_diffmicros) to explore graphs to see if they
have a tangent line at a point.
In Figure 3.5 we show the graph of f (x) = x and its tangent line at (1, 1) in a series of tighter intervals about x = 1.
As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on
the tangent line a good approximation to the values of the function for choices of x close to 1. In fact, the graph of f (x)
itself appears to be locally linear in the immediate vicinity of x = 1.
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Chapter 3 | Derivatives
217
Figure 3.5 For values of x close to 1, the graph of f (x) = x and its tangent line appear to coincide.
Formally we may define the tangent line to the graph of a function as follows.
Definition
Let f (x) be a function defined in an open interval containing a. The tangent line to f (x) at a is the line passing
through the point ⎛⎝a, f (a)⎞⎠ having slope
m tan = xlim
→a
f (x) − f (a)
x−a
(3.3)
provided this limit exists.
Equivalently, we may define the tangent line to f (x) at a to be the line passing through the point ⎛⎝a, f (a)⎞⎠ having
slope
m tan = lim
h→0
f (a + h) − f (a)
h
(3.4)
provided this limit exists.
Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the
slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend
on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find
equations of tangent lines.
218
Chapter 3 | Derivatives
Example 3.1
Finding a Tangent Line
Find the equation of the line tangent to the graph of f (x) = x 2 at x = 3.
Solution
First find the slope of the tangent line. In this example, use Equation 3.3.
m tan = lim
x→3
f (x) − f (3)
x−3
2
= lim x − 9
x→3 x−3
(x − 3)(x + 3)
= lim
= lim (x + 3) = 6
x−3
x→3
x→3
Apply the definition
Substitute f (x) = x 2 and f (3) = 9.
Factor the numerator to evaluate the limit.
Next, find a point on the tangent line. Since the line is tangent to the graph of f (x) at x = 3, it passes through
the point ⎛⎝3, f (3)⎞⎠. We have f (3) = 9, so the tangent line passes through the point (3, 9).
Using the point-slope equation of the line with the slope m = 6 and the point (3, 9), we obtain the line
y − 9 = 6(x − 3). Simplifying, we have y = 6x − 9. The graph of f (x) = x 2 and its tangent line at 3 are
shown in Figure 3.6.
Figure 3.6 The tangent line to f (x) at x = 3.
Example 3.2
The Slope of a Tangent Line Revisited
Use Equation 3.4 to find the slope of the line tangent to the graph of f (x) = x 2 at x = 3.
Solution
The steps are very similar to Example 3.1. See Equation 3.4 for the definition.
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Chapter 3 | Derivatives
219
m tan = lim
h→0
f (3 + h) − f (3)
h
Apply the definition
(3 + h) 2 − 9
h
h→0
Substitute f (3 + h) = (3 + h) 2 and f (3) = 9.
= lim
2
= lim 9 + 6h + h − 9
h
h→0
h(6 + h)
= lim (6 + h) = 6
= lim
h
h→0
h→0
Expand and simplify to evaluate the limit.
We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the
formulas can be interchanged.
Example 3.3
Finding the Equation of a Tangent Line
Find the equation of the line tangent to the graph of f (x) = 1/x at x = 2.
Solution
We can use Equation 3.3, but as we have seen, the results are the same if we use Equation 3.4.
f (x) − f (2)
x−2
1−1
x
2
= lim
x → 2x − 2
1−1
x
2 · 2x
= lim
x → 2 x − 2 2x
(2 − x)
= lim
x → 2 (x − 2)(2x)
= lim −1
x → 2 2x
= −1
4
m tan = lim
x→2
Apply the definition
Substitute f (x) = 1x and f (2) = 1 .
2
Multiply numerator and denominator by 2x to
simplify fractions.
Simplify.
Simplify using 2 − x = −1, for x ≠ 2.
x−2
Evaluate the limit.
We now know that the slope of the tangent line is − 1 . To find the equation of the tangent line, we also need a
4
1
point on the line. We know that f (2) = . Since the tangent line passes through the point (2, 1 ) we can use
2
2
the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation
y = − 1 x + 1. The graphs of f (x) = 1x and y = − 1 x + 1 are shown in Figure 3.7.
4
4
220
Chapter 3 | Derivatives
Figure 3.7 The line is tangent to f (x) at x = 2.
3.1
Find the slope of the line tangent to the graph of f (x) = x at x = 4.
The Derivative of a Function at a Point
The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications
across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in
business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative.
The process of finding a derivative is called differentiation.
Definition
Let f (x) be a function defined in an open interval containing a. The derivative of the function f (x) at a, denoted
by f ′ (a), is defined by
f ′ (a) = xlim
→a
f (x) − f (a)
x−a
(3.5)
provided this limit exists.
Alternatively, we may also define the derivative of f (x) at a as
f ′ (a) = lim
h→0
f (a + h) − f (a)
.
h
Example 3.4
Estimating a Derivative
For f (x) = x 2, use a table to estimate f ′(3) using Equation 3.5.
Solution
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(3.6)
Chapter 3 | Derivatives
221
Create a table using values of x just below 3 and just above 3.
x
x2 − 9
x−3
2.9
5.9
2.99
5.99
2.999
5.999
3.001
6.001
3.01
6.01
3.1
6.1
After examining the table, we see that a good estimate is f ′ (3) = 6.
3.2
For f (x) = x 2, use a table to estimate f ′(3) using Equation 3.6.
Example 3.5
Finding a Derivative
For f (x) = 3x 2 − 4x + 1, find f ′(2) by using Equation 3.5.
Solution
Substitute the given function and value directly into the equation.
f ′ (x) = lim
x→2
f (x) − f (2)
x−2
Apply the definition
− 4x + 1⎞⎠ − 5
x−2
x→2
(x − 2)(3x + 2)
= lim
x−2
x→2
= lim (3x + 2)
Cancel the common factor.
=8
Evaluate the limit.
= lim
x→2
⎛ 2
⎝3x
Substitute f (x) = 3x 2 − 4x + 1 and f (2) = 5.
Simplify and factor the numerator.
222
Chapter 3 | Derivatives
Example 3.6
Revisiting the Derivative
For f (x) = 3x 2 − 4x + 1, find f ′(2) by using Equation 3.6.
Solution
Using this equation, we can substitute two values of the function into the equation, and we should get the same
value as in Example 3.5.
f ′ (2) = lim
h→0
f (2 + h) − f (2)
h
Apply the definition
(3(2 + h) 2 − 4(2 + h) + 1) − 5
h
h→0
= lim
2
= lim 3h + 8h
h
h→0
h(3h + 8)
= lim
h
h→0
= lim (3h + 8)
Substitute f (2) = 5 and
f (2 + h) = 3(2 + h) 2 − 4(2 + h) + 1.
Simplify the numerator.
Factor the numerator.
Cancel the common factor.
h→0
=8
Evaluate the limit.
The results are the same whether we use Equation 3.5 or Equation 3.6.
3.3
For f (x) = x 2 + 3x + 2, find f ′ (1).
Velocities and Rates of Change
Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s(t) is the position of an object
moving along a coordinate axis, the average velocity of the object over a time interval [a, t] if t > a or [t, a] if t < a is
given by the difference quotient
v ave =
s(t) − s(a)
t−a .
(3.7)
As the values of t approach a, the values of v ave approach the value we call the instantaneous velocity at a. That is,
instantaneous velocity at a, denoted v(a), is given by
v(a) = s′ (a) = lim
t→a
s(t) − s(a)
t−a .
(3.8)
To better understand the relationship between average velocity and instantaneous velocity, see Figure 3.8. In this figure,
the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t = a whose position at time
t is given by the function s(t). The slope of the secant line (shown in green) is the average velocity of the object over the
time interval [a, t].
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Chapter 3 | Derivatives
223
Figure 3.8 The slope of the secant line is the average velocity
over the interval [a, t]. The slope of the tangent line is the
instantaneous velocity.
We can use Equation 3.5 to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by
using a table of values. We can then confirm the estimate by using Equation 3.7.
Example 3.7
Estimating Velocity
A lead weight on a spring is oscillating up and down. Its position at time t with respect to a fixed horizontal
line is given by s(t) = sin t (Figure 3.9). Use a table of values to estimate v(0). Check the estimate by using
Equation 3.5.
Figure 3.9 A lead weight suspended from a spring in vertical
oscillatory motion.
Solution
We can estimate the instantaneous velocity at t = 0 by computing a table of average velocities using values of t
approaching 0, as shown in Table 3.2.
224
Chapter 3 | Derivatives
t
sint − sin0 = sint
t
t −0
−0.1
0.998334166
−0.01
0.9999833333
−0.001
0.999999833
0.001
0.999999833
0.01
0.9999833333
0.1
0.998334166
Table 3.2
Average velocities using values of t
approaching 0
From the table we see that the average velocity over the time interval [−0.1, 0] is 0.998334166, the average
velocity over the time interval [−0.01, 0] is 0.9999833333, and so forth. Using this table of values, it appears
that a good estimate is v(0) = 1.
By using Equation 3.5, we can see that
v(0) = s′ (0) = lim sint − sin0 = lim sint
= 1.
t−0
t→0
t→0 t
Thus, in fact, v(0) = 1.
3.4
A rock is dropped from a height of 64 feet. Its height above ground at time t seconds later is given by
s(t) = −16t 2 + 64, 0 ≤ t ≤ 2. Find its instantaneous velocity 1 second after it is dropped, using Equation
3.5.
As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related
concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or
the rate of change of a function at any point along the function.
Definition
The instantaneous rate of change of a function f (x) at a value a is its derivative f ′(a).
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Chapter 3 | Derivatives
225
Example 3.8
Chapter Opener: Estimating Rate of Change of Velocity
Figure 3.10 (credit: modification of work by Codex41,
Flickr)
Reaching a top speed of 270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it
went from 0 to 60 mph in 3.05 seconds, from 0 to 100 mph in 5.88 seconds, from 0 to 200 mph in 14.51
seconds, and from 0 to 229.9 mph in 19.96 seconds. Use this data to draw a conclusion about the rate of change
of velocity (that is, its acceleration) as it approaches 229.9 mph. Does the rate at which the car is accelerating
appear to be increasing, decreasing, or constant?
Solution
First observe that 60 mph = 88 ft/s, 100 mph ≈ 146.67 ft/s, 200 mph ≈ 293.33 ft/s, and 229.9 mph
≈ 337.19 ft/s. We can summarize the information in a table.
t
v(t)
0
0
3.05
88
5.88
147.67
14.51
293.33
19.96
337.19
Table 3.3
v(t) at different values
of t
Now compute the average acceleration of the car in feet per second on intervals of the form ⎡⎣t, 19.96⎤⎦ as t
approaches 19.96, as shown in the following table.
226
Chapter 3 | Derivatives
t
v(t) − v(19.96) v(t) − 337.19
=
t − 19.96
t − 19.96
0.0
16.89
3.05
14.74
5.88
13.46
14.51
8.05
Table 3.4
Average acceleration
The rate at which the car is accelerating is decreasing as its velocity approaches 229.9 mph (337.19 ft/s).
Example 3.9
Rate of Change of Temperature
A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F at 9 p.m.,
reaches a low of 60° during the night, and rises back to 70° by 7 a.m. the next morning. Suppose that the
temperature in the house is given by T(t) = 0.4t 2 − 4t + 70 for 0 ≤ t ≤ 10, where t is the number of hours
past 9 p.m. Find the instantaneous rate of change of the temperature at midnight.
Solution
Since midnight is 3 hours past 9 p.m., we want to compute T′(3). Refer to Equation 3.5.
T′ (3) = lim
t→3
T(t) − T(3)
t−3
2
= lim 0.4t − 4t + 70 − 61.6
t−3
t→3
2
= lim 0.4t − 4t + 8.4
t−3
t→3
0.4(t − 3)(t − 7)
= lim
t−3
t→3
= lim 0.4(t − 7)
t→3
= −1.6
Apply the definition
Substitute T(t) = 0.4t 2 − 4t + 70 and
T(3) = 61.6.
Simplify.
= lim
t→3
Cancel.
0.4(t − 3)(t − 7)
t−3
Evaluate the limit.
The instantaneous rate of change of the temperature at midnight is −1.6°F per hour.
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Chapter 3 | Derivatives
227
Example 3.10
Rate of Change of Profit
A toy company can sell x electronic gaming systems at a price of p = −0.01x + 400 dollars per gaming
system. The cost of manufacturing x systems is given by C(x) = 100x + 10,000 dollars. Find the rate of change
of profit when 10,000 games are produced. Should the toy company increase or decrease production?
Solution
The profit P(x) earned by producing x gaming systems is R(x) − C(x), where R(x) is the revenue obtained
from the sale of x games. Since the company can sell x games at p = −0.01x + 400 per game,
R(x) = xp = x(−0.01x + 400) = −0.01x 2 + 400x.
Consequently,
P(x) = −0.01x 2 + 300x − 10,000.
Therefore, evaluating the rate of change of profit gives
P′ (10000) =
=
lim
x → 10000
P(x) − P(10000)
x − 10000
−0.01x 2 + 300x − 10000 − 1990000
x − 10000
x → 10000
lim
−0.01x 2 + 300x − 2000000
x − 10000
x → 10000
= 100.
=
lim
Since the rate of change of profit P′ (10,000) > 0 and P(10,000) > 0,
the company should increase
production.
3.5
A coffee shop determines that the daily profit on scones obtained by charging s dollars per scone is
P(s) = −20s 2 + 150s − 10. The coffee shop currently charges $3.25 per scone. Find P′(3.25), the rate of
change of profit when the price is $3.25 and decide whether or not the coffee shop should consider raising or
lowering its prices on scones.
228
Chapter 3 | Derivatives
3.1 EXERCISES
For the following exercises, use Equation 3.3 to find the
slope of the secant line between the values x 1 and x 2 for
each function y = f (x).
2 , a = −4
x+3
19.
f (x) =
20.
f (x) = 32 , a = 3
x
1.
f (x) = 4x + 7; x 1 = 2, x 2 = 5
2.
f (x) = 8x − 3; x 1 = −1, x 2 = 3
3.
f (x) = x 2 + 2x + 1; x 1 = 3, x 2 = 3.5
21.
f (x) = 5x + 4, a = −1
4.
f (x) = −x 2 + x + 2; x 1 = 0.5, x 2 = 1.5
22.
f (x) = −7x + 1, a = 3
5.
f (x) =
23.
f (x) = x 2 + 9x, a = 2
24.
6.
f (x) = x − 7 ; x 1 = −2, x 2 = 0
2x + 1
f (x) = 3x 2 − x + 2, a = 1
25.
f (x) = x, a = 4
26.
f (x) = x − 2, a = 6
27.
f (x) = 1x , a = 2
28.
f (x) =
29.
f (x) = 13 , a = 1
x
30.
f (x) = 1 , a = 4
x
7.
8.
9.
10.
4 ; x = 1, x = 3
2
3x − 1 1
f (x) = x; x 1 = 1, x 2 = 16
f (x) = x − 9; x 1 = 10, x 2 = 13
f (x) = x 1/3 + 1; x 1 = 0, x 2 = 8
f (x) = 6x
2/3
+ 2x
1/3
Equation 3.3.
; x 1 = 1, x 2 = 27
For the following functions,
a. use Equation 3.4 to find the slope of the tangent
line m tan = f ′ (a), and
b. find the equation of the tangent line to f at x = a.
11.
f (x) = 3 − 4x, a = 2
12.
f (x) = x + 6, a = −1
5
13.
For the following functions y = f (x), find f ′ (a) using
f (x) = x 2 + x, a = 1
2
14.
f (x) = 1 − x − x , a = 0
15.
f (x) = 7x , a = 3
16.
f (x) = x + 8, a = 1
17.
f (x) = 2 − 3x 2, a = −2
18.
f (x) = −3 , a = 4
x−1
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1 , a = −1
x−3
For the following exercises, given the function y = f (x),
a. find the slope of the secant line PQ for each point
Q⎛⎝x, f (x)⎞⎠ with x value given in the table.
b. Use the answers from a. to estimate the value of the
slope of the tangent line at P.
c. Use the answer from b. to find the equation of the
tangent line to f at point P.
Chapter 3 | Derivatives
31.
[T]
229
f (x) = x 2 + 3x + 4, P(1, 8) (Round to 6
decimal places.)
places.)
x
Slope mPQ
−0.1
(i)
−0.01
(ii)
−0.001
(iii)
(x)
−0.0001
(iv)
0.99999
(xi)
−0.00001
(v)
0.999999
(xii)
−0.000001
(vi)
x
Slope
mPQ
x
Slope
mPQ
1.1
(i)
0.9
(vii)
1.01
(ii)
0.99
(viii)
1.001
(iii)
0.999
(ix)
1.0001
(iv)
0.9999
1.00001
(v)
1.000001
(vi)
34. [T] f (x) = tan(x), P(π, 0)
32. [T] f (x) = x2+ 1 , P(0, −1)
x −1
x
33. [T] f (x) = 10e 0.5x, P(0, 10) (Round to 4 decimal
x
Slope mPQ
3.1
(i)
3.14
(ii)
3.141
(iii)
(ix)
3.1415
(iv)
−0.0001
(x)
3.14159
(v)
(v)
−0.00001
(xi)
3.141592
(vi)
(vi)
−0.000001
(xii)
Slope
mPQ
x
0.1
(i)
−0.1
(vii)
0.01
(ii)
−0.01
(viii)
0.001
(iii)
−0.001
0.0001
(iv)
0.00001
0.000001
Slope
mPQ
[T] For the following position functions y = s(t),
an
object is moving along a straight line, where t is in seconds
and s is in meters. Find
a. the simplified expression for the average velocity
from t = 2 to t = 2 + h;
average velocity between t = 2 and
t = 2 + h, where (i) h = 0.1, (ii) h = 0.01,
b. the
(iii) h = 0.001, and (iv) h = 0.0001; and
c. use the answer from a. to estimate the instantaneous
230
Chapter 3 | Derivatives
velocity at t = 2 second.
35. s(t) = 1 t + 5
3
36. s(t) = t 2 − 2t
37. s(t) = 2t 3 + 3
38. s(t) = 16
−4
t2 t
39. Use the following graph to evaluate a. f ′ (1) and b.
f ′ (6).
45. [T] The position in feet of a race car along a straight
track after t seconds is modeled by the function
s(t) = 8t 2 − 1 t 3.
16
a. Find the average velocity of the vehicle over the
following time intervals to four decimal places:
i. [4, 4.1]
ii. [4, 4.01]
iii. [4, 4.001]
iv. [4, 4.0001]
b. Use a. to draw a conclusion about the instantaneous
velocity of the vehicle at t = 4 seconds.
46. [T] The distance in feet that a ball rolls down an
incline is modeled by the function s(t) = 14t 2, where t is
seconds after the ball begins rolling.
a. Find the average velocity of the ball over the
following time intervals:
i. [5, 5.1]
ii. [5, 5.01]
iii. [5, 5.001]
iv. [5, 5.0001]
b. Use the answers from a. to draw a conclusion about
the instantaneous velocity of the ball at t = 5
seconds.
40. Use the following graph to evaluate a. f ′ (−3) and b.
f ′ (1.5).
For the following exercises, use the limit definition of
derivative to show that the derivative does not exist at
x = a for each of the given functions.
41.
f (x) = x 1/3, x = 0
42.
f (x) = x 2/3, x = 0
43.
⎧1, x < 1
f (x) = ⎨
,x=1
⎩x, x ≥ 1
44.
f (x) = |xx| , x = 0
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47. Two vehicles start out traveling side by side along
a straight road. Their position functions, shown in the
following graph, are given by s = f (t) and s = g(t),
where s is measured in feet and t is measured in seconds.
a. Which vehicle has traveled farther at t = 2
seconds?
b. What is the approximate velocity of each vehicle at
t = 3 seconds?
c. Which vehicle is traveling faster at t = 4 seconds?
d. What is true about the positions of the vehicles at
t = 4 seconds?
Chapter 3 | Derivatives
231
48. [T] The total cost C(x),
to produce
x
in hundreds of dollars,
jars of mayonnaise is given by
C(x) = 0.000003x 3 + 4x + 300.
a. Calculate the average cost per jar over the
following intervals:
i. [100, 100.1]
ii. [100, 100.01]
iii. [100, 100.001]
iv. [100, 100.0001]
b. Use the answers from a. to estimate the average
cost to produce 100 jars of mayonnaise.
49. [T] For the function f (x) = x 3 − 2x 2 − 11x + 12,
do the following.
a. Use a graphing calculator to graph f in an
appropriate viewing window.
b. Use the ZOOM feature on the calculator to
approximate the two values of x = a for which
m tan = f ′ (a) = 0.
50.
[T] For the function
f (x) =
x ,
1 + x2
following.
a. Use a graphing calculator to graph
do the
f in an
appropriate viewing window.
b. Use the ZOOM feature on the calculator to
approximate the values of x = a for which
m tan = f ′ (a) = 0.
51. Suppose that N(x) computes the number of gallons
of gas used by a vehicle traveling x miles. Suppose the
vehicle gets 30 mpg.
a. Find a mathematical expression for N(x).
b. What is N(100)? Explain the physical meaning.
c. What is N′(100) ? Explain the physical meaning.
52. [T] For the function f (x) = x 4 − 5x 2 + 4, do the
following.
a. Use a graphing calculator to graph
f in an
appropriate viewing window.
b. Use the nDeriv function, which numerically finds
the derivative, on a graphing calculator to estimate
f ′ (−2), f ′(−0.5), f ′(1.7), and f ′(2.718).
53.
[T] For the function
f (x) =
x2 ,
x +1
2
following.
a. Use a graphing calculator to graph
do the
f in an
appropriate viewing window.
b. Use the nDeriv function on a graphing calculator
to find f ′ (−4), f ′(−2), f ′(2), and f ′(4).
232
Chapter 3 | Derivatives
3.2 | The Derivative as a Function
Learning Objectives
3.2.1
3.2.2
3.2.3
3.2.4
3.2.5
Define the derivative function of a given function.
Graph a derivative function from the graph of a given function.
State the connection between derivatives and continuity.
Describe three conditions for when a function does not have a derivative.
Explain the meaning of a higher-order derivative.
As we have seen, the derivative of a function at a given point gives us the rate of change or slope of the tangent line to the
function at that point. If we differentiate a position function at a given time, we obtain the velocity at that time. It seems
reasonable to conclude that knowing the derivative of the function at every point would produce valuable information about
the behavior of the function. However, the process of finding the derivative at even a handful of values using the techniques
of the preceding section would quickly become quite tedious. In this section we define the derivative function and learn a
process for finding it.
Derivative Functions
The derivative function gives the derivative of a function at each point in the domain of the original function for which the
derivative is defined. We can formally define a derivative function as follows.
Definition
Let f be a function. The derivative function, denoted by f ′, is the function whose domain consists of those values
of x such that the following limit exists:
f ′ (x) = lim
h→0
f (x + h) − f (x)
.
h
(3.9)
A function f (x) is said to be differentiable at a if f ′(a) exists. More generally, a function is said to be differentiable
on S if it is differentiable at every point in an open set S, and a differentiable function is one in which f ′(x) exists on
its domain.
In the next few examples we use Equation 3.9 to find the derivative of a function.
Example 3.11
Finding the Derivative of a Square-Root Function
Find the derivative of f (x) = x.
Solution
Start directly with the definition of the derivative function. Use Equation 3.1.
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Chapter 3 | Derivatives
233
f ′ (x) = lim x + h − x
h
h→0
= lim x + h − x · x + h + x
h
h→0
x+h+ x
h
x + h + x)
1
= lim
h → 0 ( x + h + x)
= 1
2 x
Substitute f (x + h) = x + h and f (x) = x
f (x + h) − f (x)
into f ′ (x) = lim
.
h
h→0
Multiply numerator and denominator by
x + h + x without distributing in the
denominator.
Multiply the numerators and simplify.
= lim
h → 0 h(
Cancel the h.
Evaluate the limit.
Example 3.12
Finding the Derivative of a Quadratic Function
Find the derivative of the function f (x) = x 2 − 2x.
Solution
Follow the same procedure here, but without having to multiply by the conjugate.
((x + h) 2 − 2(x + h)) − (x 2 − 2x)
h
h→0
f ′ (x) = lim
2
2
2
= lim x + 2xh + h − 2x − 2h − x + 2x
h
h→0
f (x) = x 2 − 2x into
f (x + h) − f (x)
.
f ′ (x) = lim
h
h→0
Expand (x + h) 2 − 2(x + h).
2
= lim 2xh − 2h + h
h
h→0
h(2x − 2 + h)
= lim
h
h→0
= lim (2x − 2 + h)
Cancel the common factor of h.
= 2x − 2
Evaluate the limit.
Simplify.
Factor out h from the numerator.
h→0
3.6
Substitute f (x + h) = (x + h) 2 − 2(x + h) and
Find the derivative of f (x) = x 2.
We use a variety of different notations to express the derivative of a function. In Example 3.12 we showed that if
f (x) = x 2 − 2x, then f ′ (x) = 2x − 2. If we had expressed this function in the form y = x 2 − 2x, we could have
expressed the derivative as y′ = 2x − 2 or
dy
= 2x − 2. We could have conveyed the same information by writing
dx
d ⎛x 2 − 2x⎞ = 2x − 2. Thus, for the function y = f (x), each of the following notations represents the derivative of
⎠
dx ⎝
f (x):
234
Chapter 3 | Derivatives
f ′ (x),
In place of f ′ (a) we may also use
dy
dx
|
x=a
dy
, y′, d ⎛⎝ f (x)⎞⎠.
dx
dx
Use of the
dy
notation (called Leibniz notation) is quite common in
dx
engineering and physics. To understand this notation better, recall that the derivative of a function at a point is the limit of
the slopes of secant lines as the secant lines approach the tangent line. The slopes of these secant lines are often expressed
Δy
where Δy is the difference in the y values corresponding to the difference in the x values, which are
Δx
expressed as Δx (Figure 3.11). Thus the derivative, which can be thought of as the instantaneous rate of change of y
in the form
with respect to x, is expressed as
dy
Δy
= lim
.
dx Δx → 0 Δx
Figure 3.11 The derivative is expressed as
dy
Δy
= lim
.
dx Δx → 0 Δx
Graphing a Derivative
We have already discussed how to graph a function, so given the equation of a function or the equation of a derivative
function, we could graph it. Given both, we would expect to see a correspondence between the graphs of these two
functions, since f ′(x) gives the rate of change of a function f (x) (or slope of the tangent line to f (x)).
In Example 3.11 we found that for f (x) = x, f ′(x) = 1/2 x. If we graph these functions on the same axes, as in Figure
3.12, we can use the graphs to understand the relationship between these two functions. First, we notice that f (x) is
increasing over its entire domain, which means that the slopes of its tangent lines at all points are positive. Consequently,
we expect f ′ (x) > 0 for all values of x in its domain. Furthermore, as x increases, the slopes of the tangent lines to f (x)
are decreasing and we expect to see a corresponding decrease in f ′(x). We also observe that f (0) is undefined and that
lim f ′ (x) = +∞, corresponding to a vertical tangent to f (x) at 0.
x → 0+
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Chapter 3 | Derivatives
235
Figure 3.12 The derivative f ′(x) is positive everywhere
because the function f (x) is increasing.
In Example 3.12 we found that for f (x) = x 2 − 2x, f ′ (x) = 2x − 2. The graphs of these functions are shown in Figure
3.13. Observe that f (x) is decreasing for x < 1. For these same values of x, f ′ (x) < 0. For values of x > 1, f (x) is
increasing and f ′ (x) > 0. Also, f (x) has a horizontal tangent at x = 1 and f ′ (1) = 0.
Figure 3.13 The derivative f ′ (x) < 0 where the function
f (x) is decreasing and f ′ (x) > 0 where f (x) is increasing.
The derivative is zero where the function has a horizontal
tangent.
Example 3.13
Sketching a Derivative Using a Function
Use the following graph of f (x) to sketch a graph of f ′ (x).
236
Chapter 3 | Derivatives
Solution
The solution is shown in the following graph. Observe that f (x) is increasing and f ′ (x) > 0 on ( – 2, 3). Also,
f (x) is decreasing and f ′ (x) < 0 on (−∞, −2) and on (3, +∞). Also note that f (x) has horizontal tangents
at – 2 and 3, and f ′ (−2) = 0 and f ′ (3) = 0.
3.7
Sketch the graph of f (x) = x 2 − 4. On what interval is the graph of f ′ (x) above the x -axis?
Derivatives and Continuity
Now that we can graph a derivative, let’s examine the behavior of the graphs. First, we consider the relationship between
differentiability and continuity. We will see that if a function is differentiable at a point, it must be continuous there;
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Chapter 3 | Derivatives
237
however, a function that is continuous at a point need not be differentiable at that point. In fact, a function may be continuous
at a point and fail to be differentiable at the point for one of several reasons.
Theorem 3.1: Differentiability Implies Continuity
Let f (x) be a function and a be in its domain. If f (x) is differentiable at a, then f is continuous at a.
Proof
If f (x) is differentiable at a, then f ′(a) exists and
f ′ (a) = xlim
→a
f (x) − f (a)
x−a .
We want to show that f (x) is continuous at a by showing that xlim
f (x) = f (a). Thus,
→a
⎛
f (x) − f (a) + f (a)⎞⎠
lim f (x) = xlim
→ a⎝
x→a
⎞
⎛ f (x) − f (a)
= xlim
x − a · (x − a) + f (a)⎠
→ a⎝
f (x) − f (a) ⎞ ⎛
⎛
(x − a)⎞⎠ + xlim
f (a)
= ⎝xlim
x − a ⎠ · ⎝xlim
→a
→a
→a
Multiply and divide f (x) − f (a) by x − a.
= f ′(a) · 0 + f (a)
= f (a).
Therefore, since f (a) is defined and xlim
f (x) = f (a), we conclude that f is continuous at a.
→a
□
We have just proven that differentiability implies continuity, but now we consider whether continuity implies
differentiability. To determine an answer to this question, we examine the function f (x) = |x|. This function is continuous
everywhere; however, f ′(0) is undefined. This observation leads us to believe that continuity does not imply
differentiability. Let’s explore further. For f (x) = |x|,
f ′ (0) = lim
x→0
f (x) − f (0)
= lim |x| − |0| = lim |xx| .
x−0
x→0 x−0
x→0
This limit does not exist because
lim |xx| = −1 and lim |xx| = 1.
+
x → 0−
See Figure 3.14.
x→0
238
Chapter 3 | Derivatives
Figure 3.14 The function f (x) = |x| is continuous at 0 but
is not differentiable at 0.
Let’s consider some additional situations in which a continuous function fails to be differentiable. Consider the function
f (x) = 3 x:
3
f ′ (0) = lim x − 0 = lim 3 1 = +∞.
x→0 x−0
x→0 2
x
Thus f ′ (0) does not exist. A quick look at the graph of f (x) = 3 x clarifies the situation. The function has a vertical
tangent line at 0 (Figure 3.15).
3
Figure 3.15 The function f (x) = x has a vertical tangent at
x = 0. It is continuous at 0 but is not differentiable at 0.
⎧x sin ⎛1 ⎞ if x ≠ 0
⎝x ⎠
The function f (x) = ⎨
also has a derivative that exhibits interesting behavior at 0. We see that
⎩0 if x = 0
f ′ (0) = lim
x→0
x sin(1/x) − 0
= lim sin ⎛⎝1x ⎞⎠.
x−0
x→0
This limit does not exist, essentially because the slopes of the secant lines continuously change direction as they approach
zero (Figure 3.16).
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Chapter 3 | Derivatives
239
⎧x sin ⎛1 ⎞ if x ≠ 0
⎝x ⎠
Figure 3.16 The function f (x) = ⎨
is not
⎩0 if x = 0
differentiable at 0.
In summary:
1. We observe that if a function is not continuous, it cannot be differentiable, since every differentiable function must
be continuous. However, if a function is continuous, it may still fail to be differentiable.
2. We saw that f (x) = |x| failed to be differentiable at 0 because the limit of the slopes of the tangent lines on the
left and right were not the same. Visually, this resulted in a sharp corner on the graph of the function at 0. From
this we conclude that in order to be differentiable at a point, a function must be “smooth” at that point.
3. As we saw in the example of f (x) = 3 x, a function fails to be differentiable at a point where there is a vertical
tangent line.
⎧x sin ⎛1 ⎞ if x ≠ 0
⎝x ⎠
a function may fail to be differentiable at a point in more complicated
4. As we saw with f (x) = ⎨
⎩0 if x = 0
ways as well.
Example 3.14
A Piecewise Function that is Continuous and Differentiable
A toy company wants to design a track for a toy car that starts out along a parabolic curve and then converts
to a straight line (Figure 3.17). The function that describes the track is to have the form
⎧ 1 x 2 + bx + cif x < −10
10
f (x) = ⎨
⎩ − 14 x + 52 if x ≥ −10
where x and f (x) are in inches. For the car to move smoothly along the
track, the function f (x) must be both continuous and differentiable at −10. Find values of b and c that make
f (x) both continuous and differentiable.
240
Chapter 3 | Derivatives
Figure 3.17 For the car to move smoothly along the track, the
function must be both continuous and differentiable.
Solution
For the function to be continuous at x = −10,
lim
x → −10−
lim
x → 10−
f (x) = f (−10). Thus, since
f (x) = 1 (−10) 2 − 10b + c = 10 − 10b + c
10
and f (−10) = 5, we must have 10 − 10b + c = 5. Equivalently, we have c = 10b − 5.
For the function to be differentiable at −10,
f ′ (10) =
lim
x → −10
f (x) − f (−10)
x + 10
must exist. Since f (x) is defined using different rules on the right and the left, we must evaluate this limit from
the right and the left and then set them equal to each other:
lim
x → −10−
f (x) − f (−10)
= lim −
x + 10
x → −10
=
lim
x → −10−
1 x 2 + bx + c − 5
10
x + 10
1 x 2 + bx + (10b − 5) − 5
10
x + 10
x 2 − 100 + 10bx + 100b
10(x + 10)
(x + 10)(x − 10 + 10b)
= lim −
10(x + 10)
x → −10
= b − 2.
=
Substitute c = 10b − 5.
lim
x → −10−
Factor by grouping.
We also have
lim
x → −10+
− 14 x + 52 − 5
f (x) − f (−10)
= lim
x + 10
x + 10
x → −10+
=
lim
x → −10+
−(x + 10)
4(x + 10)
= − 1.
4
⎛ ⎞
This gives us b − 2 = − 1 . Thus b = 7 and c = 10⎝7 ⎠ − 5 = 25 .
4
3.8
4
4
2
⎧ax + b if x < 3
Find values of a and b that make f (x) = ⎨ 2
both continuous and differentiable at 3.
⎩ x if x ≥ 3
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Chapter 3 | Derivatives
241
Higher-Order Derivatives
The derivative of a function is itself a function, so we can find the derivative of a derivative. For example, the derivative
of a position function is the rate of change of position, or velocity. The derivative of velocity is the rate of change of
velocity, which is acceleration. The new function obtained by differentiating the derivative is called the second derivative.
Furthermore, we can continue to take derivatives to obtain the third derivative, fourth derivative, and so on. Collectively,
these are referred to as higher-order derivatives. The notation for the higher-order derivatives of y = f (x) can be
expressed in any of the following forms:
f ″(x), f ‴(x), f
(4)
y″(x), y‴(x), y
(4)
(x),…, y
4
n
2
3
(x),…, f
(n)
(n)
(x)
(x)
d y d y d y
d y
,
,
,…,
.
2
3
4
dy n
dx dy dy
d2 y
⎛dy ⎞
may be viewed as an attempt to express d ⎝ ⎠ more compactly.
dx dx
dx 2
It is interesting to note that the notation for
⎛ 2 ⎞
d y
d y
⎛ ⎛dy ⎞⎞
.
Analogously, d ⎝ d ⎝ ⎠⎠ = d ⎜ 2 ⎟ =
dx dx dx
dx ⎝ dx ⎠ dx 3
3
Example 3.15
Finding a Second Derivative
For f (x) = 2x 2 − 3x + 1, find f ″(x).
Solution
First find f ′(x).
⎛
⎝2(x
f ′ (x) = lim
h→0
Substitute f (x) = 2x 2 − 3x + 1
and
f (x + h) = 2(x + h) 2 − 3(x + h) + 1
f (x + h) − f (x)
.
into f ′ (x) = lim
h
h→0
+ h) 2 − 3(x + h) + 1⎞⎠ − (2x 2 − 3x + 1)
h
2
= lim 4xh + h − 3h
h
h→0
Simplify the numerator.
Factor out the h in the numerator
and cancel with the h in the
denominator.
Take the limit.
= lim (4x + h − 3)
h→0
= 4x − 3
Next, find f ″(x) by taking the derivative of f ′ (x) = 4x − 3.
f ′ (x + h) − f ′(x)
h
h→0
f ″(x) = lim
Use f ′ (x) = lim
h→0
f (x + h) − f (x)
with f ′(x) in
h
4(x + h) − 3 − (4x − 3)
h
= lim 4
place of f (x).
Substitute f ′ (x + h) = 4(x + h) − 3 and
f ′ (x) = 4x − 3.
Simplify.
=4
Take the limit.
= lim
h→0
h→0
⎛
⎝
⎞
⎠
242
Chapter 3 | Derivatives
3.9
Find f ″(x) for f (x) = x 2.
Example 3.16
Finding Acceleration
The position of a particle along a coordinate axis at time t (in seconds) is given by s(t) = 3t 2 − 4t + 1 (in
meters). Find the function that describes its acceleration at time t.
Solution
Since v(t) = s′(t) and a(t) = v′ (t) = s″(t), we begin by finding the derivative of s(t) :
s′ (t) = lim
h→0
s(t + h) − s(t)
h
3(t + h) 2 − 4(t + h) + 1 − ⎛⎝3t 2 − 4t + 1⎞⎠
h
h→0
= 6t − 4.
= lim
Next,
s′ (t + h) − s′(t)
h
6(t + h) − 4 − (6t − 4)
= lim
h
h→0
= 6.
s″(t) = lim
h→0
Thus, a = 6 m/s 2.
3.10
For s(t) = t 3, find a(t).
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Chapter 3 | Derivatives
243
3.2 EXERCISES
For the following exercises, use the definition of a
derivative to find f ′ (x).
54.
f (x) = 6
55.
f (x) = 2 − 3x
56.
f (x) = 2x + 1
7
57.
f (x) = 4x 2
58.
f (x) = 5x − x 2
59.
f (x) = 2x
60.
f (x) = x − 6
61.
f (x) = 9x
62.
f (x) = x + 1x
63.
f (x) = 1
x
For the following exercises, use the graph of y = f (x) to
sketch the graph of its derivative f ′ (x).
64.
65.
66.
244
Chapter 3 | Derivatives
67.
77.
⎧2x, x ≤ 1
f (x) = ⎨2
⎩x , x > 1
For the following graphs,
a. determine for which values of x = a the
lim f (x) exists but f is not continuous at
x→a
x = a, and
b. determine for which values of x = a the function
is continuous but not differentiable at x = a.
78.
For the following exercises, the given limit represents the
derivative of a function y = f (x) at x = a. Find f (x)
and a.
68.
69.
70.
71.
(1 + h) 2/3 − 1
h
h→0
lim
lim
h→0
lim
h→0
lim
h→0
⎡
⎣3(2
+ h) 2 + 2⎤⎦ − 14
h
cos(π + h) + 1
h
4
(2 + h) − 16
h
72.
[2(3 + h) 2 − (3 + h)] − 15
h
h→0
73.
h
lim e − 1
h→0 h
lim
For the following functions,
a. sketch the graph and
b. use the definition of a derivative to show that the
function is not differentiable at x = 1.
74.
⎧2 x, 0 ≤ x ≤ 1
f (x) = ⎨
⎩3x − 1, x > 1
75.
⎧3, x < 1
f (x) = ⎨
⎩3x, x ≥ 1
76.
⎧−x 2 + 2, x ≤ 1
f (x) = ⎨
⎩x, x > 1
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79.
Chapter 3 | Derivatives
245
80. Use the graph to evaluate a. f ′ (−0.5), b. f ′ (0), c.
f ′ (x) = lim
b.
f ′ (1), d. f ′ (2), and e. f ′ (3), if it exists.
h→0
f (x + h) − f (x)
h
P(x) denotes the population of a city at time x in
90.
years.
C(x) denotes the total amount of money (in
91.
thousands of dollars) spent on concessions by x customers
at an amusement park.
R(x) denotes the total cost (in thousands of dollars)
92.
of manufacturing x clock radios.
93. g(x) denotes the grade (in percentage points) received
on a test, given x hours of studying.
B(x) denotes the cost (in dollars) of a sociology
94.
textbook at university bookstores in the United States in x
years since 1990.
For
the
following
functions,
f ′ (x + h) − f ′(x)
f ″(x) = lim
to find f ″(x).
h
h→0
81.
f (x) = 2 − 3x
82.
f (x) = 4x 2
83.
f (x) = x + 1x
use
p(x) denotes atmospheric pressure at an altitude of x
95.
feet.
96. Sketch the graph of a function y = f (x) with all of
the following properties:
a. f ′ (x) > 0 for −2 ≤ x < 1
b.
f ′ (2) = 0
c.
f ′ (x) > 0 for x > 2
d.
f (2) = 2 and f (0) = 1
For the following exercises, use a calculator to graph f (x).
e.
Determine the function f ′ (x), then use a calculator to
f.
graph f ′ (x).
84. [T] f (x) = − 5x
2
85. [T] f (x) = 3x + 2x + 4.
86. [T] f (x) = x + 3x
87. [T] f (x) = 1
2x
88. [T] f (x) = 1 + x + 1x
89. [T] f (x) = x 3 + 1
For the following exercises, describe what the two
expressions represent in terms of each of the given
situations. Be sure to include units.
a.
f (x + h) − f (x)
h
lim f (x) = 0 and x lim
f (x) = ∞
→∞
x → −∞
f ′ (1) does not exist.
97. Suppose temperature T in degrees Fahrenheit at a
height x in feet above the ground is given by y = T(x).
a. Give a physical interpretation, with units, of T′(x).
b. If we know that T′ (1000) = −0.1, explain the
physical meaning.
98. Suppose the total profit of a company is y = P(x)
thousand dollars when x units of an item are sold.
a. What does
P(b) − P(a)
for 0 < a < b measure,
b−a
and what are the units?
b. What does P′(x) measure, and what are the units?
c. Suppose
that
P′ (30) = 5,
what
is
the
approximate change in profit if the number of items
sold increases from 30 to 31 ?
246
Chapter 3 | Derivatives
99. The graph in the following figure models the number
of people N(t) who have come down with the flu t weeks
101. [T] Construct a table of values for h′ (t) and graph
after its initial outbreak in a town with a population of
50,000 citizens.
points, estimate both the left limit and right limit and
average them.)
a. Describe what N′(t) represents and how it behaves
as t increases.
b. What does the derivative tell us about how this
town is affected by the flu outbreak?
both h(t) and h′ (t) on the same graph. (Hint: for interior
102.
[T] The best linear fit to the data is given by
H(t) = 7.229t − 4.905, where H is the height of the
rocket (in meters) and t is the time elapsed since takeoff.
From this equation, determine H′ (t). Graph H(t) with
the given data and, on a separate coordinate plane, graph
H′ (t).
103. [T] The best quadratic fit to the data is given by
G(t) = 1.429t 2 + 0.0857t − 0.1429, where G is the
height of the rocket (in meters) and t is the time elapsed
since takeoff. From this equation, determine G′ (t). Graph
G(t) with the given data and, on a separate coordinate
plane, graph G′ (t).
104.
[T] The best cubic fit to the data is given by
where
F(t) = 0.2037t 3 + 2.956t 2 − 2.705t + 0.4683,
F is the height of the rocket (in m) and t is the time
elapsed since take off. From this equation, determine
F′ (t). Graph F(t) with the given data and, on a separate
coordinate plane, graph F′ (t). Does the linear, quadratic,
For the following exercises, use the following table, which
shows the height h of the Saturn V rocket for the Apollo
11 mission t seconds after launch.
Time (seconds)
Height (meters)
0
0
1
2
2
4
3
13
4
25
5
32
100. What is the physical meaning of h′ (t) ? What are
the units?
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or cubic function fit the data best?
105. Using the best linear, quadratic, and cubic fits to
the data, determine what H″(t), G″(t) and F″(t) are. What
are the physical meanings of H″(t), G″(t) and F″(t), and
what are their units?
Chapter 3 | Derivatives
247
3.3 | Differentiation Rules
Learning Objectives
3.3.1
3.3.2
3.3.3
3.3.4
3.3.5
3.3.6
State the constant, constant multiple, and power rules.
Apply the sum and difference rules to combine derivatives.
Use the product rule for finding the derivative of a product of functions.
Use the quotient rule for finding the derivative of a quotient of functions.
Extend the power rule to functions with negative exponents.
Combine the differentiation rules to find the derivative of a polynomial or rational function.
Finding derivatives of functions by using the definition of the derivative can be a lengthy and, for certain functions, a rather
challenging process. For example, previously we found that d ( x) = 1
dx
2 x
by using a process that involved multiplying an
⎛ ⎞
expression by a conjugate prior to evaluating a limit. The process that we could use to evaluate d ⎝3 x⎠ using the definition,
dx
while similar, is more complicated. In this section, we develop rules for finding derivatives that allow us to bypass this
process. We begin with the basics.
The Basic Rules
The functions f (x) = c and g(x) = x n where n is a positive integer are the building blocks from which all polynomials
and rational functions are constructed. To find derivatives of polynomials and rational functions efficiently without resorting
to the limit definition of the derivative, we must first develop formulas for differentiating these basic functions.
The Constant Rule
We first apply the limit definition of the derivative to find the derivative of the constant function, f (x) = c. For this
function, both f (x) = c and f (x + h) = c, so we obtain the following result:
f (x + h) − f (x)
h
= lim c − c
h→0 h
= lim 0
h → 0h
= lim 0 = 0.
f ′ (x) = lim
h→0
h→0
The rule for differentiating constant functions is called the constant rule. It states that the derivative of a constant function
is zero; that is, since a constant function is a horizontal line, the slope, or the rate of change, of a constant function is 0. We
restate this rule in the following theorem.
Theorem 3.2: The Constant Rule
Let c be a constant.
If f (x) = c, then f ′(c) = 0.
Alternatively, we may express this rule as
d (c) = 0.
dx
Example 3.17
248
Chapter 3 | Derivatives
Applying the Constant Rule
Find the derivative of f (x) = 8.
Solution
This is just a one-step application of the rule:
f ′(8) = 0.
3.11
Find the derivative of g(x) = −3.
The Power Rule
We have shown that
d ⎛x 2⎞ = 2x and d ⎛x 1/2⎞ = 1 x −1/2.
⎠
2
dx ⎝ ⎠
dx ⎝
d (x n). We continue our
dx
examination of derivative formulas by differentiating power functions of the form f (x) = x n where n is a positive integer.
At this point, you might see a pattern beginning to develop for derivatives of the form
We develop formulas for derivatives of this type of function in stages, beginning with positive integer powers. Before stating
and proving the general rule for derivatives of functions of this form, we take a look at a specific case, d (x 3). As we go
dx
through this derivation, pay special attention to the portion of the expression in boldface, as the technique used in this case
is essentially the same as the technique used to prove the general case.
Example 3.18
Differentiating x3
Find d ⎛⎝x 3⎞⎠.
dx
Solution
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Chapter 3 | Derivatives
249
3
3
d ⎛x 3⎞ = lim (x + h) − x
⎠
⎝
dx
h
h→0
3
2
Notice that the fir t term in the expansion of
2
3
= lim x + 3x h + 3xh + h − x
h
h→0
3
In this step the x 3 terms have been cancelled,
leaving only terms containing h.
3
2
2
= lim 3x h + 3xh + h
h
h→0
h(3x 2 + 3xh + h 2)
h
h→0
= lim
Factor out the common factor of h.
After cancelling the common factor of h, the
= lim (3x 2 + 3xh + h 2)
only term not containing h is 3x 2.
h→0
= 3x 2
3.12
(x + h) 3 is x 3 and the second term is 3x 2 h. All
other terms contain powers of h that are two or
greater.
Let h go to 0.
Find d ⎛⎝x 4⎞⎠.
dx
As we shall see, the procedure for finding the derivative of the general form f (x) = x n is very similar. Although it is often
unwise to draw general conclusions from specific examples, we note that when we differentiate f (x) = x 3, the power on
x becomes the coefficient of x 2 in the derivative and the power on x in the derivative decreases by 1. The following
theorem states that the power rule holds for all positive integer powers of x. We will eventually extend this result to
negative integer powers. Later, we will see that this rule may also be extended first to rational powers of x and then to
arbitrary powers of x. Be aware, however, that this rule does not apply to functions in which a constant is raised to a
variable power, such as f (x) = 3 x.
Theorem 3.3: The Power Rule
Let n be a positive integer. If f (x) = x n, then
f ′ (x) = nx n − 1.
Alternatively, we may express this rule as
d x n = nx n − 1.
dx
Proof
For f (x) = x n where n is a positive integer, we have
(x + h) n − x n
.
h
h→0
⎛n⎞
⎛n⎞
Since (x + h) n = x n + nx n − 1 h + ⎝ ⎠ x n − 2 h 2 + ⎝ ⎠ x n − 3 h 3 + … + nxh n − 1 + h n,
3
2
f ′ (x) = lim
we see that
250
Chapter 3 | Derivatives
⎛n⎞
⎛n⎞
(x + h) n − x n = nx n − 1 h + ⎝ ⎠ x n − 2 h 2 + ⎝ ⎠ x n − 3 h 3 + … + nxh n − 1 + h n.
3
2
Next, divide both sides by h:
⎛n⎞
⎛n⎞
n−1
h + ⎝ ⎠ x n − 2 h 2 + ⎝ ⎠ x n − 3 h 3 + … + nxh n − 1 + h n
(x + h) n − x n nx
3
2
=
.
h
h
Thus,
(x + h) n − x n
⎛n⎞
⎛n⎞
= nx n − 1 + ⎝ ⎠ x n − 2 h + ⎝ ⎠ x n − 3 h 2 + … + nxh n − 2 + h n − 1.
3
2
h
Finally,
⎛n⎞
⎞
⎛
⎛n⎞
f ′ (x) = lim ⎝nx n − 1 + ⎝ ⎠ x n − 2 h + ⎝ ⎠ x n − 3 h 2 + … + nxh n − 1 + h n⎠
3
2
h→0
= nx n − 1.
□
Example 3.19
Applying the Power Rule
Find the derivative of the function f (x) = x 10 by applying the power rule.
Solution
Using the power rule with n = 10, we obtain
f ′(x) = 10x 10 − 1 = 10x 9.
3.13
Find the derivative of f (x) = x 7.
The Sum, Difference, and Constant Multiple Rules
We find our next differentiation rules by looking at derivatives of sums, differences, and constant multiples of functions.
Just as when we work with functions, there are rules that make it easier to find derivatives of functions that we add, subtract,
or multiply by a constant. These rules are summarized in the following theorem.
Theorem 3.4: Sum, Difference, and Constant Multiple Rules
Let f (x) and g(x) be differentiable functions and k be a constant. Then each of the following equations holds.
Sum Rule. The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f
and the derivative of g.
d ⎛⎝ f (x) + g(x)⎞⎠ = d ⎛⎝ f (x)⎞⎠ + d ⎛⎝g(x)⎞⎠;
dx
dx
dx
that is,
for j(x) = f (x) + g(x), j′ (x) = f ′ (x) + g′(x).
Difference Rule. The derivative of the difference of a function f and a function g is the same as the difference of the
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Chapter 3 | Derivatives
251
derivative of f and the derivative of g:
d ⎛⎝ f (x) − g(x)⎞⎠ = d ⎛⎝ f (x)⎞⎠ − d ⎛⎝g(x)⎞⎠;
dx
dx
dx
that is,
for j(x) = f (x) − g(x), j′ (x) = f ′ (x) − g′(x).
Constant Multiple Rule. The derivative of a constant c multiplied by a function f is the same as the constant multiplied
by the derivative:
d ⎛⎝k f (x)⎞⎠ = k d ⎛⎝ f (x)⎞⎠;
dx
dx
that is,
for j(x) = k f (x), j′ (x) = k f ′(x).
Proof
We provide only the proof of the sum rule here. The rest follow in a similar manner.
For differentiable functions f (x) and g(x), we set j(x) = f (x) + g(x). Using the limit definition of the derivative we
have
j′ (x) = lim
h→0
j(x + h) − j(x)
.
h
By substituting j(x + h) = f (x + h) + g(x + h) and j(x) = f (x) + g(x), we obtain
j′(x) = lim
h→0
⎛
⎝
f (x + h) + g(x + h)⎞⎠ − ⎛⎝ f (x) + g(x)⎞⎠
.
h
Rearranging and regrouping the terms, we have
⎛ f (x + h) − f (x) g(x + h) − g(x) ⎞
j′(x) = lim ⎝
+
⎠.
h
h
h→0
We now apply the sum law for limits and the definition of the derivative to obtain
⎛ f (x + h) − f (x) ⎞
⎛g(x + h) − g(x) ⎞
j′(x) = lim ⎝
⎠ = f ′ (x) + g′ (x).
⎠ + hlim
h
h
h→0
→ 0⎝
□
Example 3.20
Applying the Constant Multiple Rule
Find the derivative of g(x) = 3x 2 and compare it to the derivative of f (x) = x 2.
Solution
We use the power rule directly:
g′ (x) = d ⎛⎝3x 2⎞⎠ = 3 d ⎛⎝x 2⎞⎠ = 3(2x) = 6x.
dx
dx
Since f (x) = x 2 has derivative f ′ (x) = 2x, we see that the derivative of g(x) is 3 times the derivative of
252
Chapter 3 | Derivatives
f (x). This relationship is illustrated in Figure 3.18.
Figure 3.18 The derivative of g(x) is 3 times the derivative of f (x).
Example 3.21
Applying Basic Derivative Rules
Find the derivative of f (x) = 2x 5 + 7.
Solution
We begin by applying the rule for differentiating the sum of two functions, followed by the rules for
differentiating constant multiples of functions and the rule for differentiating powers. To better understand the
sequence in which the differentiation rules are applied, we use Leibniz notation throughout the solution:
f ′ (x) = d ⎛⎝2x 5 + 7⎞⎠
dx
= d ⎛⎝2x 5⎞⎠ + d (7)
dx
dx
⎛ 5⎞
d
= 2 ⎝x ⎠ + d (7)
dx
dx
= 2⎛⎝5x 4⎞⎠ + 0
= 10x 4.
3.14
Apply the sum rule.
Apply the constant multiple rule.
Apply the power rule and the constant rule.
Simplify.
Find the derivative of f (x) = 2x 3 − 6x 2 + 3.
Example 3.22
Finding the Equation of a Tangent Line
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Chapter 3 | Derivatives
253
Find the equation of the line tangent to the graph of f (x) = x 2 − 4x + 6 at x = 1.
Solution
To find the equation of the tangent line, we need a point and a slope. To find the point, compute
f (1) = 1 2 − 4(1) + 6 = 3.
This gives us the point (1, 3). Since the slope of the tangent line at 1 is f ′ (1), we must first find f ′ (x). Using
the definition of a derivative, we have
f ′ (x) = 2x − 4
so the slope of the tangent line is f ′ (1) = −2. Using the point-slope formula, we see that the equation of the
tangent line is
y − 3 = −2(x − 1).
Putting the equation of the line in slope-intercept form, we obtain
y = −2x + 5.
3.15
Find the equation of the line tangent to the graph of f (x) = 3x 2 − 11 at x = 2. Use the point-slope
form.
The Product Rule
Now that we have examined the basic rules, we can begin looking at some of the more advanced rules. The first one
examines the derivative of the product of two functions. Although it might be tempting to assume that the derivative of
the product is the product of the derivatives, similar to the sum and difference rules, the product rule does not follow this
pattern. To see why we cannot use this pattern, consider the function f (x) = x 2, whose derivative is f ′ (x) = 2x and not
d (x) · d (x) = 1 · 1 = 1.
dx
dx
Theorem 3.5: Product Rule
Let f (x) and g(x) be differentiable functions. Then
d ⎛⎝ f (x)g(x)⎞⎠ = d ⎛⎝ f (x)⎞⎠ · g(x) + d ⎛⎝g(x)⎞⎠ · f (x).
dx
dx
dx
That is,
if j(x) = f (x)g(x), then j′ (x) = f ′ (x)g(x) + g′ (x) f (x).
This means that the derivative of a product of two functions is the derivative of the first function times the second
function plus the derivative of the second function times the first function.
Proof
We begin by assuming that f (x) and g(x) are differentiable functions. At a key point in this proof we need to use the
fact that, since g(x) is differentiable, it is also continuous. In particular, we use the fact that since g(x) is continuous,
lim g(x + h) = g(x).
h→0
254
Chapter 3 | Derivatives
By applying the limit definition of the derivative to j(x) = f (x)g(x), we obtain
j′ (x) = lim
h→0
f (x + h)g(x + h) − f (x)g(x)
.
h
By adding and subtracting f (x)g(x + h) in the numerator, we have
j′ (x) = lim
h→0
f (x + h)g(x + h) − f (x)g(x + h) + f (x)g(x + h) − f (x)g(x)
.
h
After breaking apart this quotient and applying the sum law for limits, the derivative becomes
⎛ f (x + h)g(x + h) − f (x)g(x + h) ⎞
⎛ f (x)g(x + h) − f (x)g(x) ⎞
j′ (x) = lim ⎝
⎠ + hlim
⎠.
h
h
h→0
→ 0⎝
Rearranging, we obtain
⎛ f (x + h) − f (x)
⎞
⎞
⎛g(x + h) − g(x)
j′ (x) = lim ⎝
· g(x + h)⎠ + lim ⎝
· f (x)⎠.
h
h
h→0
h→0
By using the continuity of g(x), the definition of the derivatives of f (x) and g(x), and applying the limit laws, we arrive
at the product rule,
j′ (x) = f ′ (x)g(x) + g′ (x) f (x).
□
Example 3.23
Applying the Product Rule to Constant Functions
For j(x) = f (x)g(x), use the product rule to find j′(2) if f (2) = 3, f ′ (2) = −4, g(2) = 1, and g′ (2) = 6.
Solution
Since j(x) = f (x)g(x), j′ (x) = f ′ (x)g(x) + g′ (x) f (x), and hence
j′ (2) = f ′ (2)g(2) + g′ (2) f (2) = (−4)(1) + (6)(3) = 14.
Example 3.24
Applying the Product Rule to Binomials
For j(x) = (x 2 + 2)(3x 3 − 5x), find j′(x) by applying the product rule. Check the result by first finding the
product and then differentiating.
Solution
If we set f (x) = x 2 + 2 and g(x) = 3x 3 − 5x, then f ′ (x) = 2x and g′ (x) = 9x 2 − 5. Thus,
j′ (x) = f ′ (x)g(x) + g′ (x) f (x) = (2x)⎛⎝3x 3 − 5x⎞⎠ + (9x 2 − 5)(x 2 + 2).
Simplifying, we have
j′ (x) = 15x 4 + 3x 2 − 10.
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Chapter 3 | Derivatives
255
To check, we see that j(x) = 3x 5 + x 3 − 10x and, consequently, j′ (x) = 15x 4 + 3x 2 − 10.
3.16
Use the product rule to obtain the derivative of j(x) = 2x 5 ⎛⎝4x 2 + x⎞⎠.
The Quotient Rule
Having developed and practiced the product rule, we now consider differentiating quotients of functions. As we see in the
following theorem, the derivative of the quotient is not the quotient of the derivatives; rather, it is the derivative of the
function in the numerator times the function in the denominator minus the derivative of the function in the denominator
times the function in the numerator, all divided by the square of the function in the denominator. In order to better grasp
why we cannot simply take the quotient of the derivatives, keep in mind that
d ⎛x 2⎞ = 2x, not
dx ⎝ ⎠
d ⎛x 3⎞
dx ⎝ ⎠
d (x)
dx
2
= 3x = 3x 2.
1
Theorem 3.6: The Quotient Rule
Let f (x) and g(x) be differentiable functions. Then
d ⎛ f (x) ⎞ =
dx ⎝ g(x) ⎠
d ( f (x)) · g(x) − d (g(x)) ·
dx
dx
2
(g(x))
f (x)
.
That is,
if j(x) =
f (x)
f ′ (x)g(x) − g′ (x) f (x)
, then j′ (x) =
.
g(x)
(g(x)) 2
The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this
new rule for finding derivatives in the next example.
Example 3.25
Applying the Quotient Rule
2
Use the quotient rule to find the derivative of k(x) = 5x .
4x + 3
Solution
Let f (x) = 5x 2 and g(x) = 4x + 3. Thus, f ′ (x) = 10x and g′ (x) = 4. Substituting into the quotient rule, we
have
k′ (x) =
f ′ (x)g(x) − g′ (x) f (x) 10x(4x + 3) − 4(5x 2)
=
.
(g(x)) 2
(4x + 3) 2
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Chapter 3 | Derivatives
Simplifying, we obtain
2
k′ (x) = 20x + 30x
.
(4x + 3) 2
3.17
Find the derivative of h(x) = 3x + 1 .
4x − 3
It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form x k where k
is a negative integer.
Theorem 3.7: Extended Power Rule
If k is a negative integer, then
d ⎛x k⎞ = kx k − 1.
dx ⎝ ⎠
Proof
If k is a negative integer, we may set n = −k, so that n is a positive integer with k = −n. Since for each positive integer
n, x −n = 1n , we may now apply the quotient rule by setting f (x) = 1 and g(x) = x n. In this case, f ′ (x) = 0 and
x
g′ (x) = nx n − 1. Thus,
⎛ n − 1⎞
n
⎠
d (x −n) = 0(x ) − 1⎝nx
.
n 2
d
(x )
Simplifying, we see that
d (x −n) = −nx n − 1 = −nx (n − 1) − 2n = −nx −n − 1.
d
x 2n
Finally, observe that since k = −n, by substituting we have
d ⎛x k⎞ = kx k − 1.
dx ⎝ ⎠
□
Example 3.26
Using the Extended Power Rule
Find d ⎛⎝x −4⎞⎠.
dx
Solution
By applying the extended power rule with k = −4, we obtain
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Chapter 3 | Derivatives
257
d ⎛x −4⎞ = −4x −4 − 1 = −4x −5.
dx ⎝ ⎠
Example 3.27
Using the Extended Power Rule and the Constant Multiple Rule
Use the extended power rule and the constant multiple rule to find f (x) = 62 .
x
Solution
It may seem tempting to use the quotient rule to find this derivative, and it would certainly not be incorrect to do
so. However, it is far easier to differentiate this function by first rewriting it as f (x) = 6x −2.
⎛ ⎞
f ′ (x) = d 62 = d ⎛⎝6x −2⎞⎠
dx ⎝x ⎠ dx
= 6 d (x −2)
dx
= 6(−2x −3)
= −12x −3
3.18
Rewrite 62 as 6x −2.
x
Apply the constant multiple rule.
Use the extended power rule to diffe entiate x −2.
Simplify.
Find the derivative of g(x) = 17 using the extended power rule.
x
Combining Differentiation Rules
As we have seen throughout the examples in this section, it seldom happens that we are called on to apply just one
differentiation rule to find the derivative of a given function. At this point, by combining the differentiation rules, we may
find the derivatives of any polynomial or rational function. Later on we will encounter more complex combinations of
differentiation rules. A good rule of thumb to use when applying several rules is to apply the rules in reverse of the order in
which we would evaluate the function.
Example 3.28
Combining Differentiation Rules
For k(x) = 3h(x) + x 2 g(x), find k′(x).
Solution
Finding this derivative requires the sum rule, the constant multiple rule, and the product rule.
258
Chapter 3 | Derivatives
k′ (x) = d ⎛⎝3h(x) + x 2 g(x)⎞⎠ = d ⎛⎝3h(x)⎞⎠ + d ⎛⎝x 2 g(x)⎞⎠
dx
dx
dx
⎛
⎞
= 3 d ⎛⎝h(x)⎞⎠ + ⎝ d ⎛⎝x 2⎞⎠g(x) + d ⎛⎝g(x)⎞⎠x 2⎠
dx
dx
dx
= 3h′ (x) + 2xg(x) + g′(x)x 2
Apply the sum rule.
Apply the constant multiple rule to
diffe entiate 3h(x) and the product
rule to diffe entiate x 2 g(x).
Example 3.29
Extending the Product Rule
For k(x) = f (x)g(x)h(x), express k′ (x) in terms of f (x), g(x), h(x), and their derivatives.
Solution
We can think of the function k(x) as the product of the function f (x)g(x) and the function h(x). That is,
k(x) = ⎛⎝ f (x)g(x)⎞⎠ · h(x). Thus,
k′(x) = d ⎛⎝ f (x)g(x)⎞⎠ · h(x) + d ⎛⎝h(x)⎞⎠ · ⎛⎝ f (x)g(x)⎞⎠
dx
dx
= ⎛⎝ f ′ (x)g(x) + g′ (x) f (x)h⎞⎠(x) + h′ (x) f (x)g(x)
= f ′ (x)g(x)h(x) + f (x)g′ (x)h(x) + f (x)g(x)h′ (x).
Apply the product rule to the product
of f (x)g(x) and h(x).
Apply the product rule to f (x)g(x).
Simplify.
Example 3.30
Combining the Quotient Rule and the Product Rule
For h(x) =
2x 3 k(x)
, find h′(x).
3x + 2
Solution
This procedure is typical for finding the derivative of a rational function.
h′ (x) =
=
=
d ⎛2x 3 k(x)⎞ · (3x + 2) − d (3x + 2) · ⎛2x 3 k(x)⎞
⎠
⎠
⎝
dx
dx ⎝
2
⎛ 2
⎝6x k(x)
(3x + 2)
+ k′ (x) · 2x 3⎞⎠(3x + 2) − 3⎛⎝2x 3 k(x)⎞⎠
(3x + 2) 2
−6x 3 k(x) + 18x 3 k(x) + 12x 2 k(x) + 6x 4 k′ (x) + 4x 3 k′ (x)
(3x + 2) 2
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Apply the quotient rule.
Apply the product rule to fin
d ⎛2x 3 k(x)⎞. Use d (3x + 2) = 3.
⎠
dx
dx ⎝
Simplify.
Chapter 3 | Derivatives
3.19
259
Find d ⎛⎝3 f (x) − 2g(x)⎞⎠.
dx
Example 3.31
Determining Where a Function Has a Horizontal Tangent
Determine the values of x for which f (x) = x 3 − 7x 2 + 8x + 1 has a horizontal tangent line.
Solution
To find the values of x for which f (x) has a horizontal tangent line, we must solve f ′ (x) = 0. Since
f ′ (x) = 3x 2 − 14x + 8 = (3x − 2)(x − 4),
we must solve (3x − 2)(x − 4) = 0. Thus we see that the function has horizontal tangent lines at x = 2 and
3
x = 4 as shown in the following graph.
Figure 3.19 This function has horizontal tangent lines at x =
2/3 and x = 4.
Example 3.32
Finding a Velocity
The position of an object on a coordinate axis at time t is given by s(t) = 2 t . What is the initial velocity of
t +1
the object?
Solution
Since the initial velocity is v(0) = s′ (0), begin by finding s′(t) by applying the quotient rule:
s′ (t) =
1⎛⎝t 2 + 1⎞⎠ − 2t(t)
⎛ 2
⎝t
2
⎞
+ 1⎠
2
= 1 − t 2.
⎛ 2
⎞
⎝t + 1⎠
260
Chapter 3 | Derivatives
After evaluating, we see that v(0) = 1.
3.20
Find the values of x for which the line tangent to the graph of f (x) = 4x 2 − 3x + 2 has a tangent line
parallel to the line y = 2x + 3.
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Chapter 3 | Derivatives
261
Formula One Grandstands
Formula One car races can be very exciting to watch and attract a lot of spectators. Formula One track designers have
to ensure sufficient grandstand space is available around the track to accommodate these viewers. However, car racing
can be dangerous, and safety considerations are paramount. The grandstands must be placed where spectators will not
be in danger should a driver lose control of a car (Figure 3.20).
Figure 3.20 The grandstand next to a straightaway of the Circuit de Barcelona-Catalunya race track, located where
the spectators are not in danger.
Safety is especially a concern on turns. If a driver does not slow down enough before entering the turn, the car may
slide off the racetrack. Normally, this just results in a wider turn, which slows the driver down. But if the driver loses
control completely, the car may fly off the track entirely, on a path tangent to the curve of the racetrack.
Suppose you are designing a new Formula One track. One section of the track can be modeled by the function
f (x) = x 3 + 3x + x (Figure 3.21). The current plan calls for grandstands to be built along the first straightaway
and around a portion of the first curve. The plans call for the front corner of the grandstand to be located at the point
(−1.9, 2.8). We want to determine whether this location puts the spectators in danger if a driver loses control of the
car.
262
Chapter 3 | Derivatives
Figure 3.21 (a) One section of the racetrack can be modeled by the function f (x) = x 3 + 3x + x. (b) The
front corner of the grandstand is located at (−1.9, 2.8).
1. Physicists have determined that drivers are most likely to lose control of their cars as they are coming into a
turn, at the point where the slope of the tangent line is 1. Find the (x, y) coordinates of this point near the turn.
2. Find the equation of the tangent line to the curve at this point.
3. To determine whether the spectators are in danger in this scenario, find the x-coordinate of the point where the
tangent line crosses the line y = 2.8. Is this point safely to the right of the grandstand? Or are the spectators
in danger?
4. What if a driver loses control earlier than the physicists project? Suppose a driver loses control at the point
(−2.5, 0.625). What is the slope of the tangent line at this point?
5. If a driver loses control as described in part 4, are the spectators safe?
6. Should you proceed with the current design for the grandstand, or should the grandstands be moved?
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Chapter 3 | Derivatives
263
3.3 EXERCISES
For the following exercises, find f ′(x) for each function.
106.
f (x) = x 7 + 10
107.
f (x) = 5x 3 − x + 1
108.
f (x) = 4x 2 − 7x
4
122. h(x) = 4 f (x) +
g(x)
7
123. h(x) = x 3 f (x)
2
124. h(x) =
f (x)g(x)
2
125. h(x) =
3 f (x)
g(x) + 2
109.
f (x) = 8x + 9x − 1
110.
f (x) = x 4 + 2x
111.
⎛
⎞
f (x) = 3x⎝18x 4 + 13 ⎠
x+1
112.
f (x) = (x + 2)⎛⎝2x 2 − 3⎞⎠
x
1
2
3
4
113.
⎞
⎛
f (x) = x 2 22 + 53
⎝x
x ⎠
f(x)
3
5
−2
0
114.
3
2
f (x) = x + 2x − 4
3
g(x)
2
3
−4
6
115.
3
+1
f (x) = 4x − 2x
2
x
f′(x)
−1
7
8
−3
116.
2
f (x) = x 2 + 4
x −4
g′(x)
4
1
2
9
117.
f (x) =
For the following exercises, assume that f (x) and g(x)
x+9
x 2 − 7x + 1
are both differentiable functions with values as given in
the following table. Use the following table to calculate the
following derivatives.
126. Find h′(1) if h(x) = x f (x) + 4g(x).
f (x)
.
g(x)
For the following exercises, find the equation of the tangent
line T(x) to the graph of the given function at the indicated
127. Find h′ (2) if h(x) =
point. Use a graphing calculator to graph the function and
the tangent line.
128. Find h′ (3) if h(x) = 2x + f (x)g(x).
118. [T] y = 3x 2 + 4x + 1 at (0, 1)
g(x)
129. Find h′ (4) if h(x) = 1x +
.
119. [T] y = 2 x + 1 at (4, 5)
120. [T] y = 2x
x−1
at (−1, 1)
121. [T] y = 2x − 32 at (1, −1)
x
For the following exercises, assume that f (x) and g(x)
are both differentiable functions for all x. Find the
derivative of each of the functions h(x).
f (x)
For the following exercises, use the following figure to find
the indicated derivatives, if they exist.
264
Chapter 3 | Derivatives
140. Find the point on the graph of f (x) = x 3 such that
the tangent line at that point has an x intercept of 6.
141. Find the equation of the line passing through the
point P(3, 3) and tangent to the graph of f (x) =
142.
points on the graph of
f (x) = x + x − x − 1 for which the slope of the tangent
130. Let h(x) = f (x) + g(x). Find
a.
h′ (1),
b.
h′ (3), and
c.
h′ (4).
131. Let h(x) = f (x)g(x). Find
a.
h′ (1),
b.
h′ (3), and
c.
h′ (4).
132. Let h(x) =
f (x)
. Find
g(x)
a.
h′ (1),
b.
h′ (3), and
c.
h′ (4).
For the following exercises,
a. evaluate f ′ (a), and
b. graph the function f (x) and the tangent line at
x = a.
133. [T] f (x) = 2x 3 + 3x − x 2, a = 2
134. [T] f (x) = 1x − x 2, a = 1
135. [T] f (x) = x 2 − x 12 + 3x + 2, a = 0
136. [T] f (x) = 1x − x 2/3, a = −1
137. Find the equation of the tangent line to the graph of
f (x) = 2x 3 + 4x 2 − 5x − 3 at x = −1.
138. Find the equation of the tangent line to the graph of
f (x) = x 2 + 4x − 10 at x = 8.
139. Find the equation of the tangent line to the graph of
f (x) = (3x − x 2)(3 − x − x 2) at x = 1.
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Determine
6 .
x−1
3
all
2
line is
a. horizontal
b. −1.
143.
Find
a
quadratic
polynomial
f (1) = 5, f ′ (1) = 3 and f ″(1) = −6.
such
that
144. A car driving along a freeway with traffic has
traveled s(t) = t 3 − 6t 2 + 9t meters in t seconds.
a. Determine the time in seconds when the velocity of
the car is 0.
b. Determine the acceleration of the car when the
velocity is 0.
145. [T] A herring swimming along a straight line has
2
traveled s(t) = 2 t
feet in t seconds. Determine the
t +2
velocity of the herring when it has traveled 3 seconds.
146. The population in millions of arctic flounder in the
Atlantic Ocean is modeled by the function
P(t) = 8t 2+ 3 , where t is measured in years.
0.2t + 1
a. Determine the initial flounder population.
b. Determine P′ (10) and briefly interpret the result.
147. [T] The concentration of antibiotic in the
bloodstream t hours after being injected is given by the
2
+t,
function C(t) = 2t
3
t + 50
where C is measured in
milligrams per liter of blood.
a. Find the rate of change of C(t).
b. Determine the rate of change for t = 8, 12, 24,
and 36.
c. Briefly describe what seems to be occurring as the
number of hours increases.
148.
A book publisher has a cost function given by
3
+ 3 , where x is the number of copies of
C(x) = x + 2x
x2
a book in thousands and C is the cost, per book, measured
in dollars. Evaluate C′ (2) and explain its meaning.
Chapter 3 | Derivatives
265
149. [T] According to Newton’s law of universal
gravitation, the force F between two bodies of constant
mass m 1 and m 2 is given by the formula F =
Gm 1 m 2
,
d2
where G is the gravitational constant and d is the distance
between the bodies.
a. Suppose that G, m 1, and m 2 are constants. Find
the rate of change of force F with respect to
distance d.
b. Find the rate of change of force F with
gravitational
constant
G = 6.67 × 10 −11
Nm 2 /kg 2, on two bodies 10 meters apart, each
with a mass of 1000 kilograms.
266
Chapter 3 | Derivatives
3.4 | Derivatives as Rates of Change
Learning Objectives
3.4.1 Determine a new value of a quantity from the old value and the amount of change.
3.4.2 Calculate the average rate of change and explain how it differs from the instantaneous rate
of change.
3.4.3 Apply rates of change to displacement, velocity, and acceleration of an object moving along
a straight line.
3.4.4 Predict the future population from the present value and the population growth rate.
3.4.5 Use derivatives to calculate marginal cost and revenue in a business situation.
In this section we look at some applications of the derivative by focusing on the interpretation of the derivative as the rate of
change of a function. These applications include acceleration and velocity in physics, population growth rates in biology,
and marginal functions in economics.
Amount of Change Formula
One application for derivatives is to estimate an unknown value of a function at a point by using a known value of a
function at some given point together with its rate of change at the given point. If f (x) is a function defined on an interval
⎡
⎣
a, a + h⎤⎦, then the amount of change of f (x) over the interval is the change in the y values of the function over that
interval and is given by
f (a + h) − f (a).
The average rate of change of the function f over that same interval is the ratio of the amount of change over that interval
to the corresponding change in the x values. It is given by
f (a + h) − f (a)
.
h
As we already know, the instantaneous rate of change of f (x) at a is its derivative
f ′ (a) = lim
h→0
For small enough values of h, f ′ (a) ≈
f (a + h) − f (a)
.
h
f (a + h) − f (a)
. We can then solve for f (a + h) to get the amount of change
h
formula:
f (a + h) ≈ f (a) + f ′(a)h.
(3.10)
We can use this formula if we know only f (a) and f ′(a) and wish to estimate the value of f (a + h). For example, we
may use the current population of a city and the rate at which it is growing to estimate its population in the near future. As
we can see in Figure 3.22, we are approximating f (a + h) by the y coordinate at a + h on the line tangent to f (x) at
x = a. Observe that the accuracy of this estimate depends on the value of h as well as the value of f ′ (a).
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Chapter 3 | Derivatives
267
Figure 3.22 The new value of a changed quantity equals the
original value plus the rate of change times the interval of
change: f (a + h) ≈ f (a) + f ′ (a)h.
Here is an interesting demonstration (http://www.openstaxcollege.org/l/20_chainrule) of rate of change.
Example 3.33
Estimating the Value of a Function
If f (3) = 2 and f ′ (3) = 5, estimate f (3.2).
Solution
Begin by finding h. We have h = 3.2 − 3 = 0.2. Thus,
f (3.2) = f (3 + 0.2) ≈ f (3) + (0.2) f ′ (3) = 2 + 0.2(5) = 3.
3.21
Given f (10) = −5 and f ′ (10) = 6, estimate f (10.1).
Motion along a Line
Another use for the derivative is to analyze motion along a line. We have described velocity as the rate of change of position.
If we take the derivative of the velocity, we can find the acceleration, or the rate of change of velocity. It is also important to
introduce the idea of speed, which is the magnitude of velocity. Thus, we can state the following mathematical definitions.
Definition
Let s(t) be a function giving the position of an object at time t.
The velocity of the object at time t is given by v(t) = s′ (t).
The speed of the object at time t is given by |v(t)|.
The acceleration of the object at t is given by a(t) = v′ (t) = s″(t).
268
Chapter 3 | Derivatives
Example 3.34
Comparing Instantaneous Velocity and Average Velocity
A ball is dropped from a height of 64 feet. Its height above ground (in feet) t seconds later is given by
s(t) = −16t 2 + 64.
a. What is the instantaneous velocity of the ball when it hits the ground?
b. What is the average velocity during its fall?
Solution
The first thing to do is determine how long it takes the ball to reach the ground. To do this, set s(t) = 0. Solving
−16t 2 + 64 = 0, we get t = 2, so it take 2 seconds for the ball to reach the ground.
a. The instantaneous velocity of the ball as it strikes the ground is v(2). Since v(t) = s′ (t) = −32t, we
obtain v(t) = −64 ft/s.
b. The average velocity of the ball during its fall is
v ave =
s(2) − s(0) 0 − 64
=
= −32 ft/s.
2−0
2
Example 3.35
Interpreting the Relationship between v(t) and a(t)
A particle moves along a coordinate axis in the positive direction to the right. Its position at time t is given by
s(t) = t 3 − 4t + 2. Find v(1) and a(1) and use these values to answer the following questions.
a. Is the particle moving from left to right or from right to left at time t = 1 ?
b. Is the particle speeding up or slowing down at time t = 1 ?
Solution
Begin by finding v(t) and a(t).
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Chapter 3 | Derivatives
269
and a(t) = v′ (t) = s″(t) = 6t.
Evaluating these functions at t = 1, we obtain v(1) = −1 and a(1) = 6.
a. Because v(1) < 0, the particle is moving from right to left.
b. Because v(1) < 0 and a(1) > 0, velocity and acceleration are acting in opposite directions. In other
words, the particle is being accelerated in the direction opposite the direction in which it is traveling,
causing |v(t)| to decrease. The particle is slowing down.
Example 3.36
Position and Velocity
The position of a particle moving along a coordinate axis is given by s(t) = t 3 − 9t 2 + 24t + 4, t ≥ 0.
a. Find v(t).
b. At what time(s) is the particle at rest?
c. On what time intervals is the particle moving from left to right? From right to left?
d. Use the information obtained to sketch the path of the particle along a coordinate axis.
Solution
a. The velocity is the derivative of the position function:
v(t) = s′ (t) = 3t 2 − 18t + 24.
b. The particle is at rest when v(t) = 0, so set 3t 2 − 18t + 24 = 0. Factoring the left-hand side of the
equation produces 3(t − 2)(t − 4) = 0. Solving, we find that the particle is at rest at t = 2 and t = 4.
c. The particle is moving from left to right when v(t) > 0 and from right to left when v(t) < 0. Figure
3.23 gives the analysis of the sign of v(t) for t ≥ 0, but it does not represent the axis along which the
particle is moving.
Figure 3.23 The sign of v(t) determines the direction of the
particle.
Since 3t 2 − 18t + 24 > 0 on [0, 2) ∪ (2, +∞), the particle is moving from left to right on these
intervals.
Since 3t 2 − 18t + 24 < 0 on (2, 4), the particle is moving from right to left on this interval.
d. Before we can sketch the graph of the particle, we need to know its position at the time it starts
moving ⎛⎝t = 0) and at the times that it changes direction (t = 2, 4). We have s(0) = 4, s(2) = 24, and
s(4) = 20. This means that the particle begins on the coordinate axis at 4 and changes direction at 0 and
270
Chapter 3 | Derivatives
20 on the coordinate axis. The path of the particle is shown on a coordinate axis in Figure 3.24.
Figure 3.24 The path of the particle can be determined by
analyzing v(t).
3.22
A particle moves along a coordinate axis. Its position at time t is given by s(t) = t 2 − 5t + 1. Is the
particle moving from right to left or from left to right at time t = 3 ?
Population Change
In addition to analyzing velocity, speed, acceleration, and position, we can use derivatives to analyze various types of
populations, including those as diverse as bacteria colonies and cities. We can use a current population, together with a
growth rate, to estimate the size of a population in the future. The population growth rate is the rate of change of a population
and consequently can be represented by the derivative of the size of the population.
Definition
If P(t) is the number of entities present in a population, then the population growth rate of P(t) is defined to be
P′ (t).
Example 3.37
Estimating a Population
The population of a city is tripling every 5 years. If its current population is 10,000, what will be its approximate
population 2 years from now?
Solution
Let P(t) be the population (in thousands) t years from now. Thus, we know that P(0) = 10 and based on the
information, we anticipate P(5) = 30. Now estimate P′ (0), the current growth rate, using
P′ (0) ≈
P(5) − P(0) 30 − 10
=
= 4.
5−0
5
By applying Equation 3.10 to P(t), we can estimate the population 2 years from now by writing
P(2) ≈ P(0) + (2)P′ (0) ≈ 10 + 2(4) = 18;
thus, in 2 years the population will be 18,000.
3.23
The current population of a mosquito colony is known to be 3,000; that is, P(0) = 3,000. If
P′ (0) = 100, estimate the size of the population in 3 days, where t is measured in days.
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Chapter 3 | Derivatives
271
Changes in Cost and Revenue
In addition to analyzing motion along a line and population growth, derivatives are useful in analyzing changes in cost,
revenue, and profit. The concept of a marginal function is common in the fields of business and economics and implies the
use of derivatives. The marginal cost is the derivative of the cost function. The marginal revenue is the derivative of the
revenue function. The marginal profit is the derivative of the profit function, which is based on the cost function and the
revenue function.
Definition
If C(x) is the cost of producing x items, then the marginal cost MC(x) is MC(x) = C′ (x).
If R(x) is the revenue obtained from selling x items, then the marginal revenue MR(x) is MR(x) = R′ (x).
If P(x) = R(x) − C(x) is the profit obtained from selling x items, then the marginal profit MP(x) is defined to be
MP(x) = P′ (x) = MR(x) − MC(x) = R′ (x) − C′ (x).
We can roughly approximate
MC(x) = C′ (x) = lim
h→0
C(x + h) − C(x)
h
by choosing an appropriate value for h. Since x represents objects, a reasonable and small value for h is 1. Thus, by
substituting h = 1, we get the approximation MC(x) = C′ (x) ≈ C(x + 1) − C(x). Consequently, C′ (x) for a given
value of x can be thought of as the change in cost associated with producing one additional item. In a similar way,
MR(x) = R′ (x) approximates the revenue obtained by selling one additional item, and MP(x) = P′ (x) approximates the
profit obtained by producing and selling one additional item.
Example 3.38
Applying Marginal Revenue
Assume that the number of barbeque dinners that can be sold, x, can be related to the price charged, p, by the
equation p(x) = 9 − 0.03x, 0 ≤ x ≤ 300.
In this case, the revenue in dollars obtained by selling x barbeque dinners is given by
R(x) = xp(x) = x(9 − 0.03x) = −0.03x 2 + 9x for 0 ≤ x ≤ 300.
Use the marginal revenue function to estimate the revenue obtained from selling the 101st barbeque dinner.
Compare this to the actual revenue obtained from the sale of this dinner.
Solution
First, find the marginal revenue function: MR(x) = R′ (x) = −0.06x + 9.
Next, use R′ (100) to approximate R(101) − R(100), the revenue obtained from the sale of the 101st dinner.
Since R′ (100) = 3, the revenue obtained from the sale of the 101st dinner is approximately $3.
The actual revenue obtained from the sale of the 101st dinner is
R(101) − R(100) = 602.97 − 600 = 2.97, or $2.97.
The marginal revenue is a fairly good estimate in this case and has the advantage of being easy to compute.
272
Chapter 3 | Derivatives
3.24
Suppose that the profit obtained from the sale of
2
x
fish-fry dinners is given by
P(x) = −0.03x + 8x − 50. Use the marginal profit function to estimate the profit from the sale of the 101st
fish-fry dinner.
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Chapter 3 | Derivatives
273
3.4 EXERCISES
For the following exercises, the given functions represent
the position of a particle traveling along a horizontal line.
a. Find the velocity and acceleration functions.
b. Determine the time intervals when the object is
slowing down or speeding up.
3
2
150. s(t) = 2t − 3t − 12t + 8
151. s(t) = 2t 3 − 15t 2 + 36t − 10
152. s(t) =
t
1 + t2
153. A rocket is fired vertically upward from the ground.
The distance s in feet that the rocket travels from the
ground after t seconds is given by s(t) = −16t 2 + 560t.
157. A potato is launched vertically upward with an initial
velocity of 100 ft/s from a potato gun at the top of an
85-foot-tall building. The distance in feet that the potato
travels from the ground after t seconds is given by
s(t) = −16t 2 + 100t + 85.
a. Find the velocity of the potato after 0.5 s and
5.75 s.
b. Find the speed of the potato at 0.5 s and 5.75 s.
c. Determine when the potato reaches its maximum
height.
d. Find the acceleration of the potato at 0.5 s and 1.5
s.
e. Determine how long the potato is in the air.
f. Determine the velocity of the potato upon hitting
the ground.
158.
The position function s(t) = t 3 − 8t gives the
a. Find the velocity of the rocket 3 seconds after being
fired.
b. Find the acceleration of the rocket 3 seconds after
being fired.
position in miles of a freight train where east is the positive
direction and t is measured in hours.
a. Determine the direction the train is traveling when
154. A ball is thrown downward with a speed of 8 ft/
s from the top of a 64-foot-tall building. After t seconds,
its height above the ground is given by
b. Determine the direction the train is traveling when
s(t) = −16t 2 − 8t + 64.
a. Determine how long it takes for the ball to hit the
ground.
b. Determine the velocity of the ball when it hits the
ground.
s(t) = 0.
a(t) = 0.
c. Determine the time intervals when the train is
slowing down or speeding up.
159. The following graph shows the position y = s(t) of
an object moving along a straight line.
155. The position function s(t) = t 2 − 3t − 4 represents
the position of the back of a car backing out of a driveway
and then driving in a straight line, where s is in feet and
t is in seconds. In this case, s(t) = 0 represents the time
at which the back of the car is at the garage door, so
s(0) = −4 is the starting position of the car, 4 feet inside
the garage.
a. Determine the velocity of the car when s(t) = 0.
b. Determine the velocity of the car when s(t) = 14.
156. The position of a hummingbird flying along a straight
line in t seconds is given by s(t) = 3t 3 − 7t meters.
a. Determine the velocity of the bird at t = 1 sec.
b. Determine the acceleration of the bird at t = 1 sec.
c. Determine the acceleration of the bird when the
velocity equals 0.
a. Use the graph of the position function to determine
the time intervals when the velocity is positive,
negative, or zero.
b. Sketch the graph of the velocity function.
c. Use the graph of the velocity function to determine
the time intervals when the acceleration is positive,
negative, or zero.
d. Determine the time intervals when the object is
speeding up or slowing down.
274
Chapter 3 | Derivatives
160. The cost function, in dollars, of a company that
manufactures
food
processors
is
given
by
2
C(x) = 200 + 7x + x , where x is the number of food
7
processors manufactured.
a. Find the marginal cost function.
b. Find the marginal cost of manufacturing 12 food
processors.
c. Find the actual cost of manufacturing the thirteenth
food processor.
161. The price p (in dollars) and the demand x for a
certain digital clock radio is given by the price–demand
function p = 10 − 0.001x.
164. A small town in Ohio commissioned an actuarial
firm to conduct a study that modeled the rate of change
of the town’s population. The study found that the town’s
population (measured in thousands of people) can be
modeled by the function P(t) = − 1 t 3 + 64t + 3000,
3
where t is measured in years.
a. Find the rate of change function P′ (t) of the
population function.
b. Find P′ (1), P′ (2), P′ (3), and P′ (4). Interpret
what the results mean for the town.
c. Find P″(1), P″(2), P″(3), and P″(4). Interpret
what the results mean for the town’s population.
a. Find the revenue function R(x).
165. [T] A culture of bacteria grows in number according
b. Find the marginal revenue function.
c. Find the marginal revenue at x = 2000 and 5000.
to the function N(t) = 3000 1 + 2 4t
⎝ t + 100 ⎠, where t is
measured in hours.
a. Find the rate of change of the number of bacteria.
b. Find N′ (0), N′ (10), N′ (20), and N′ (30).
162. [T] A profit is earned when revenue exceeds cost.
Suppose the profit function for a skateboard manufacturer
is given by P(x) = 30x − 0.3x 2 − 250, where x is the
number of skateboards sold.
a. Find the exact profit from the sale of the thirtieth
skateboard.
b. Find the marginal profit function and use it to
estimate the profit from the sale of the thirtieth
skateboard.
163. [T] In general, the profit function is the difference
between
the
revenue
and
cost
functions:
P(x) = R(x) − C(x). Suppose the price-demand and cost
functions for the production of cordless drills is given
respectively
by
p = 143 − 0.03x
and
C(x) = 75,000 + 65x,
where
x is the number of
cordless drills that are sold at a price of p dollars per drill
and C(x) is the cost of producing x cordless drills.
a. Find the marginal cost function.
b. Find the revenue and marginal revenue functions.
c. Find R′(1000) and R′(4000). Interpret the
results.
d. Find the profit and marginal profit functions.
e. Find P′(1000) and P′(4000). Interpret the
results.
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⎛
c. Interpret the results in (b).
d. Find N″(0), N″(10), N″(20),
⎞
and
N″(30).
Interpret what the answers imply about the bacteria
population growth.
166. The centripetal force of an object of mass m is given
2
by F(r) = mv
r , where v is the speed of rotation and r
is the distance from the center of rotation.
a. Find the rate of change of centripetal force with
respect to the distance from the center of rotation.
b. Find the rate of change of centripetal force of an
object with mass 1000 kilograms, velocity of 13.89
m/s, and a distance from the center of rotation of
200 meters.
The following questions concern the population (in
millions) of London by decade in the 19th century, which is
listed in the following table.
Chapter 3 | Derivatives
275
Years since 1800
Population (millions)
Time after dropping (s)
Position (m)
1
0.8795
0
0
11
1.040
1
−1
21
1.264
2
−2
31
1.516
3
−5
41
1.661
4
−7
51
2.000
5
−14
61
2.634
71
3.272
81
3.911
91
4.422
Table 3.5 Population of London Source:
http://en.wikipedia.org/wiki/
Demographics_of_London.
167. [T]
a. Using a calculator or a computer program, find the
best-fit linear function to measure the population.
b. Find the derivative of the equation in a. and explain
its physical meaning.
c. Find the second derivative of the equation and
explain its physical meaning.
168. [T]
a. Using a calculator or a computer program, find the
best-fit quadratic curve through the data.
b. Find the derivative of the equation and explain its
physical meaning.
c. Find the second derivative of the equation and
explain its physical meaning.
For the following exercises, consider an astronaut on a
large planet in another galaxy. To learn more about the
composition of this planet, the astronaut drops an electronic
sensor into a deep trench. The sensor transmits its vertical
position every second in relation to the astronaut’s position.
The summary of the falling sensor data is displayed in the
following table.
169. [T]
a. Using a calculator or computer program, find the
best-fit quadratic curve to the data.
b. Find the derivative of the position function and
explain its physical meaning.
c. Find the second derivative of the position function
and explain its physical meaning.
170. [T]
a. Using a calculator or computer program, find the
best-fit cubic curve to the data.
b. Find the derivative of the position function and
explain its physical meaning.
c. Find the second derivative of the position function
and explain its physical meaning.
d. Using the result from c. explain why a cubic
function is not a good choice for this problem.
The following problems deal with the Holling type I, II,
and III equations. These equations describe the ecological
event of growth of a predator population given the amount
of prey available for consumption.
171. [T] The Holling type I equation is described by
f (x) = ax, where x is the amount of prey available and
a > 0 is the rate at which the predator meets the prey for
consumption.
a. Graph the Holling type I equation, given a = 0.5.
b. Determine the first derivative of the Holling type I
equation and explain physically what the derivative
implies.
c. Determine the second derivative of the Holling type
I equation and explain physically what the
derivative implies.
d. Using the interpretations from b. and c. explain
why the Holling type I equation may not be
realistic.
276
Chapter 3 | Derivatives
172. [T] The Holling type II equation is described by
f (x) = n ax
+ x , where x is the amount of prey available
and a > 0 is the maximum consumption rate of the
predator.
a. Graph the Holling type II equation given a = 0.5
and n = 5. What are the differences between the
Holling type I and II equations?
b. Take the first derivative of the Holling type II
equation and interpret the physical meaning of the
derivative.
c. Show that f (n) = 1 a and interpret the meaning of
2
the parameter n.
d. Find and interpret the meaning of the second
derivative. What makes the Holling type II function
more realistic than the Holling type I function?
173. [T] The Holling type III equation is described by
2
ax , where x is the amount of prey available
n2 + x2
and a > 0 is the maximum consumption rate of the
f (x) =
predator.
a. Graph the Holling type III equation given a = 0.5
and n = 5. What are the differences between the
Holling type II and III equations?
b. Take the first derivative of the Holling type III
equation and interpret the physical meaning of the
derivative.
c. Find and interpret the meaning of the second
derivative (it may help to graph the second
derivative).
d. What additional ecological phenomena does the
Holling type III function describe compared with
the Holling type II function?
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174. [T] The populations of the snowshoe hare (in
thousands) and the lynx (in hundreds) collected over 7
years from 1937 to 1943 are shown in the following table.
The snowshoe hare is the primary prey of the lynx.
Population of snowshoe
hare (thousands)
Population of
lynx (hundreds)
20
10
55
15
65
55
95
60
Table 3.6 Snowshoe Hare and Lynx
Populations Source: http://www.biotopics.co.uk/
newgcse/predatorprey.html.
a. Graph the data points and determine which
Holling-type function fits the data best.
b. Using the meanings of the parameters a and n,
determine values for those parameters by
examining a graph of the data. Recall that n
measures what prey value results in the halfmaximum of the predator value.
c. Plot the resulting Holling-type I, II, and III
functions on top of the data. Was the result from
part a. correct?
Chapter 3 | Derivatives
277
3.5 | Derivatives of Trigonometric Functions
Learning Objectives
3.5.1 Find the derivatives of the sine and cosine function.
3.5.2 Find the derivatives of the standard trigonometric functions.
3.5.3 Calculate the higher-order derivatives of the sine and cosine.
One of the most important types of motion in physics is simple harmonic motion, which is associated with such systems
as an object with mass oscillating on a spring. Simple harmonic motion can be described by using either sine or cosine
functions. In this section we expand our knowledge of derivative formulas to include derivatives of these and other
trigonometric functions. We begin with the derivatives of the sine and cosine functions and then use them to obtain formulas
for the derivatives of the remaining four trigonometric functions. Being able to calculate the derivatives of the sine and
cosine functions will enable us to find the velocity and acceleration of simple harmonic motion.
Derivatives of the Sine and Cosine Functions
We begin our exploration of the derivative for the sine function by using the formula to make a reasonable guess at its
derivative. Recall that for a function f (x),
f ′ (x) = lim
h→0
Consequently, for values of h very close to 0, f ′ (x) ≈
f (x + h) − f (x)
.
h
f (x + h) − f (x)
. We see that by using h = 0.01,
h
d (sin x) ≈ sin(x + 0.01) − sin x
0.01
dx
By setting D(x) =
sin(x + 0.01) − sin x
and using a graphing utility, we can get a graph of an approximation to the
0.01
derivative of sin x (Figure 3.25).
Figure 3.25 The graph of the function D(x) looks a lot like a
cosine curve.
Upon inspection, the graph of D(x) appears to be very close to the graph of the cosine function. Indeed, we will show that
d (sin x) = cos x.
dx
If we were to follow the same steps to approximate the derivative of the cosine function, we would find that
278
Chapter 3 | Derivatives
d (cos x) = −sinx.
dx
Theorem 3.8: The Derivatives of sin x and cos x
The derivative of the sine function is the cosine and the derivative of the cosine function is the negative sine.
d (sin x) = cos x
dx
d (cos x) = −sin x
dx
(3.11)
(3.12)
Proof
Because the proofs for d (sin x) = cos x and d (cos x) = −sin x use similar techniques, we provide only the proof for
dx
dx
d (sin x) = cos x. Before beginning, recall two important trigonometric limits we learned in Introduction to Limits:
dx
lim sinh = 1 and lim coshh − 1 = 0.
h
h
h→0
h→0
The graphs of y =
(sinh)
(cosh − 1)
and y =
are shown in Figure 3.26.
h
h
Figure 3.26 These graphs show two important limits needed to establish the derivative formulas for the
sine and cosine functions.
We also recall the following trigonometric identity for the sine of the sum of two angles:
sin(x + h) = sin x cosh + cos x sinh.
Now that we have gathered all the necessary equations and identities, we proceed with the proof.
d sin x = lim sin(x + h) − sin x
dx
h
h→0
sin
x
cosh
+ cos x sinh − sin x
= lim
h
h→0
⎛sin x cosh − sin x cos x sinh ⎞
= lim ⎝
+
⎠
h
h
h→0
⎛
⎛cos xh − 1 ⎞
⎛sinh ⎞⎞
= lim ⎝sin x⎝
⎠ + cos x⎝ h ⎠⎠
h
h→0
= sin x(0) + cos x(1)
= cos x
Apply the definition of he derivative.
Use trig identity for the sine of the sum of two angles.
Regroup.
Factor out sin x and cos x.
Apply trig limit formulas.
Simplify.
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Chapter 3 | Derivatives
279
□
Figure 3.27 shows the relationship between the graph of f (x) = sin x and its derivative f ′ (x) = cos x. Notice that at the
points where f (x) = sin x has a horizontal tangent, its derivative f ′ (x) = cos x takes on the value zero. We also see that
where f (x) = sin x is increasing, f ′ (x) = cos x > 0 and where f (x) = sin x is decreasing, f ′ (x) = cos x < 0.
Figure 3.27 Where f (x) has a maximum or a minimum,
f ′(x) = 0 that is, f ′(x) = 0 where f (x) has a horizontal
tangent. These points are noted with dots on the graphs.
Example 3.39
Differentiating a Function Containing sin x
Find the derivative of f (x) = 5x 3 sin x.
Solution
Using the product rule, we have
f ′(x) = d ⎛⎝5x 3⎞⎠ · sin x + d (sin x) · 5x 3
dx
dx
2
= 15x · sin x + cos x · 5x 3.
After simplifying, we obtain
f ′ (x) = 15x 2 sin x + 5x 3 cos x.
3.25
Find the derivative of f (x) = sin x cos x.
Example 3.40
Finding the Derivative of a Function Containing cos x
280
Chapter 3 | Derivatives
Find the derivative of g(x) = cos2x .
4x
Solution
By applying the quotient rule, we have
g′ (x) =
(−sin x)4x 2 − 8x(cos x)
⎛ 2⎞
⎝4x ⎠
2
.
Simplifying, we obtain
2
g′ (x) = −4x sin x −4 8x cos x
16x
−x
sin
x
− 2cos x .
=
4x 3
3.26
x .
Find the derivative of f (x) = cos
x
Example 3.41
An Application to Velocity
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t) = 2sint − t
for 0 ≤ t ≤ 2π. At what times is the particle at rest?
Solution
To determine when the particle is at rest, set s′ (t) = v(t) = 0. Begin by finding s′ (t). We obtain
s′ (t) = 2cost − 1,
so we must solve
2cost − 1 = 0 for 0 ≤ t ≤ 2π.
The solutions to this equation are t = π and t = 5π . Thus the particle is at rest at times t = π and t = 5π .
3
3.27
3
3
3
A particle moves along a coordinate axis. Its position at time t is given by s(t) = 3t + 2cost for
0 ≤ t ≤ 2π. At what times is the particle at rest?
Derivatives of Other Trigonometric Functions
Since the remaining four trigonometric functions may be expressed as quotients involving sine, cosine, or both, we can use
the quotient rule to find formulas for their derivatives.
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Chapter 3 | Derivatives
281
Example 3.42
The Derivative of the Tangent Function
Find the derivative of f (x) = tan x.
Solution
Start by expressing tan x as the quotient of sin x and cos x :
sin x .
f (x) = tan x = cos
x
Now apply the quotient rule to obtain
f ′ (x) =
cos x cos x − (−sin x)sin x
.
(cos x) 2
Simplifying, we obtain
2
2
f ′ (x) = cos x +2 sin x .
cos x
Recognizing that cos 2 x + sin 2 x = 1, by the Pythagorean theorem, we now have
f ′ (x) =
1 .
cos 2 x
1 to obtain
Finally, use the identity sec x = cos
x
f ′ (x) = sec 2 x.
3.28
Find the derivative of f (x) = cot x.
The derivatives of the remaining trigonometric functions may be obtained by using similar techniques. We provide these
formulas in the following theorem.
Theorem 3.9: Derivatives of tanx, cotx, sec x, and csc x
The derivatives of the remaining trigonometric functions are as follows:
d (tan x) = sec 2 x
dx
d (cot x) = −csc 2 x
dx
d (sec x) = sec x tan x
dx
d (csc x) = −csc x cot x.
dx
Example 3.43
(3.13)
(3.14)
(3.15)
(3.16)
282
Chapter 3 | Derivatives
Finding the Equation of a Tangent Line
Find the equation of a line tangent to the graph of f (x) = cot x at x = π .
4
Solution
To find the equation of the tangent line, we need a point and a slope at that point. To find the point, compute
⎛ ⎞
f ⎝π ⎠ = cot π = 1.
4
4
⎛
⎞
Thus the tangent line passes through the point ⎝π , 1⎠. Next, find the slope by finding the derivative of
4
f (x) = cot x and evaluating it at π :
4
⎛ ⎞
⎛ ⎞
f ′ (x) = −csc 2 x and f ′ ⎝π ⎠ = −csc 2 ⎝π ⎠ = −2.
4
4
Using the point-slope equation of the line, we obtain
⎛
⎞
y − 1 = −2⎝x − π ⎠
4
or equivalently,
y = −2x + 1 + π .
2
Example 3.44
Finding the Derivative of Trigonometric Functions
Find the derivative of f (x) = csc x + x tan x.
Solution
To find this derivative, we must use both the sum rule and the product rule. Using the sum rule, we find
f ′ (x) = d (csc x) + d (x tan x).
dx
dx
In the first term, d (csc x) = −csc x cot x, and by applying the product rule to the second term we obtain
dx
d (x tan x) = (1)(tan x) + (sec 2 x)(x).
dx
Therefore, we have
f ′ (x) = −csc x cot x + tan x + x sec 2 x.
3.29
Find the derivative of f (x) = 2tan x − 3cot x.
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Chapter 3 | Derivatives
3.30
283
Find the slope of the line tangent to the graph of f (x) = tan x at x = π .
6
Higher-Order Derivatives
The higher-order derivatives of sin x and cos x follow a repeating pattern. By following the pattern, we can find any
higher-order derivative of sin x and cos x.
Example 3.45
Finding Higher-Order Derivatives of y = sinx
Find the first four derivatives of y = sin x.
Solution
Each step in the chain is straightforward:
y
dy
dx
d2 y
dx 2
d3 y
dx 3
d4 y
dx 4
= sin x
= cos x
= −sin x
= −cos x
= sin x.
Analysis
Once we recognize the pattern of derivatives, we can find any higher-order derivative by determining the step in
the pattern to which it corresponds. For example, every fourth derivative of sin x equals sin x, so
d 4 (sin x) = d 8 (sin x) =
dx 4
dx 8
d 5 (sin x) = d 9 (sin x) =
dx 9
dx 5
3.31
For y = cos x, find
d 12 (sin x) = … = d 4n (sin x) = sin x
dx 12
dx 4n
d 13 (sin x) = … = d 4n + 1 (sin x) = cos x.
dx 13
dx 4n + 1
d4 y
.
dx 4
Example 3.46
Using the Pattern for Higher-Order Derivatives of y = sinx
74
Find d 74 (sin x).
dx
284
Chapter 3 | Derivatives
Solution
We can see right away that for the 74th derivative of sin x, 74 = 4(18) + 2, so
d 74 (sin x) = d 72 + 2 (sin x) = d 2 (sin x) = −sin x.
dx 2
dx 74
dx 72 + 2
3.32
59
For y = sin x, find d 59 (sin x).
dx
Example 3.47
An Application to Acceleration
A particle moves along a coordinate axis in such a way that its position at time t is given by s(t) = 2 − sint.
Find v(π/4) and a(π/4). Compare these values and decide whether the particle is speeding up or slowing down.
Solution
First find v(t) = s′ (t):
v(t) = s′ (t) = −cost.
Thus,
⎛ ⎞
v⎝π ⎠ = − 1 .
4
2
Next, find a(t) = v′(t). Thus, a(t) = v′ (t) = sint and we have
⎛ ⎞
a⎝π ⎠ = 1 .
4
2
⎛ ⎞
⎛ ⎞
Since v⎝π ⎠ = − 1 < 0 and a⎝π ⎠ = 1 > 0, we see that velocity and acceleration are acting in opposite
4
2
4
2
directions; that is, the object is being accelerated in the direction opposite to the direction in which it is travelling.
Consequently, the particle is slowing down.
3.33
A block attached to a spring is moving vertically. Its position at time t is given by s(t) = 2sint. Find
⎛ ⎞
⎛ ⎞
v⎝5π ⎠ and a⎝5π ⎠. Compare these values and decide whether the block is speeding up or slowing down.
6
6
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Chapter 3 | Derivatives
285
3.5 EXERCISES
For the following exercises, find
dy
for the given
dx
193. y = x − 1 sin x
functions.
2
175. y = x 2 − sec x + 1
194. y = 1x + tan x
176. y = 3csc x + 5x
195. y = 2csc x
177. y = x 2 cot x
196. y = sec 2 x
178. y = x − x 3 sin x
Find all x values on the graph of
f (x) = −3sin x cos x where the tangent line is horizontal.
197.
x
179. y = sec
x
198. Find all x values on the graph of f (x) = x − 2cos x
180. y = sin x tan x
for 0 < x < 2π where the tangent line has slope 2.
181. y = (x + cos x)(1 − sin x)
182. y =
192. y = sin x cos x
199. Let f (x) = cot x. Determine the points on the graph
of f for 0 < x < 2π where the tangent line(s) is (are)
tan x
1 − sec x
parallel to the line y = −2x.
183. y = 1 − cot x
200. [T] A mass on a spring bounces up and down in
simple harmonic motion, modeled by the function
s(t) = −6cost where s is measured in inches and t is
1 + cot x
184. y = cos x(1 + csc x)
For the following exercises, find the equation of the tangent
line to each of the given functions at the indicated values
of x. Then use a calculator to graph both the function and
the tangent line to ensure the equation for the tangent line
is correct.
185. [T] f (x) = −sin x, x = 0
2
t ≥ 0.
b. Find the velocity function.
c. Sketch one period of the velocity function for
188. [T] f (x) = sec x, x = π
4
t ≥ 0.
2
d. Determine the times when the velocity is 0 over one
period.
e. Find the acceleration function.
f. Sketch one period of the acceleration function for
189. [T] f (x) = x − tan x x = 0
190. [T] f (x) = 5cot x x = π
4
191. y = x sin x − cos x
the constants a and b such that when the velocity is 3 cm/
s, s = 0 and t = 0.
cm at t seconds after the jump.
a. Sketch one period of the position function for
187. [T] f (x) = 1 + cos x, x = 3π
2
functions.
201. Let the position of a swinging pendulum in simple
harmonic motion be given by s(t) = acost + bsint. Find
202. After a diver jumps off a diving board, the edge of
the board oscillates with position given by s(t) = −5cost
186. [T] f (x) = csc x, x = π
For the following exercises, find
measured in seconds. Find the rate at which the spring is
oscillating at t = 5 s.
2
d y
for the given
dx 2
t ≥ 0.
286
Chapter 3 | Derivatives
203. The number of hamburgers sold at a fast-food
restaurant in Pasadena, California, is given by
y = 10 + 5sin x where y is the number of hamburgers
sold and x represents the number of hours after the
restaurant opened at 11 a.m. until 11 p.m., when the store
closes. Find y′ and determine the intervals where the
number of burgers being sold is increasing.
204. [T] The amount of rainfall per month in Phoenix,
Arizona, can be approximated by y(t) = 0.5 + 0.3cost,
where t is months since January. Find y′ and use a
calculator to determine the intervals where the amount of
rain falling is decreasing.
For the following exercises, use the quotient rule to derive
the given equations.
205.
d (cot x) = −csc 2 x
dx
206.
d (sec x) = sec x tan x
dx
207.
d (csc x) = −csc x cot x
dx
208.
Use the definition of derivative and the identity
to prove that
cos(x + h) = cos x cosh − sin x sinh
d(cos x)
= −sin x.
dx
For the following exercises, find the requested higher-order
derivative for the given functions.
209.
d3 y
of y = 3cos x
dx 3
210.
d2 y
of y = 3sin x + x 2 cos x
2
dx
211.
d4 y
of y = 5cos x
dx 4
212.
d2 y
of y = sec x + cot x
dx 2
213.
d3 y
of y = x 10 − sec x
dx 3
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Chapter 3 | Derivatives
287
3.6 | The Chain Rule
Learning Objectives
3.6.1 State the chain rule for the composition of two functions.
3.6.2 Apply the chain rule together with the power rule.
3.6.3 Apply the chain rule and the product/quotient rules correctly in combination when both are
necessary.
3.6.4 Recognize the chain rule for a composition of three or more functions.
3.6.5 Describe the proof of the chain rule.
We have seen the techniques for differentiating basic functions (x n, sin x, cos x, etc.) as well as sums, differences,
products, quotients, and constant multiples of these functions. However, these techniques do not allow us to differentiate
compositions of functions, such as h(x) = sin⎛⎝x 3⎞⎠ or k(x) = 3x 2 + 1. In this section, we study the rule for finding the
derivative of the composition of two or more functions.
Deriving the Chain Rule
When we have a function that is a composition of two or more functions, we could use all of the techniques we have already
learned to differentiate it. However, using all of those techniques to break down a function into simpler parts that we are
able to differentiate can get cumbersome. Instead, we use the chain rule, which states that the derivative of a composite
function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
To put this rule into context, let’s take a look at an example: h(x) = sin ⎛⎝x 3⎞⎠. We can think of the derivative of this function
with respect to x as the rate of change of sin⎛⎝x 3⎞⎠ relative to the change in x. Consequently, we want to know how sin⎛⎝x 3⎞⎠
changes as x changes. We can think of this event as a chain reaction: As x changes, x 3 changes, which leads to a change
in sin ⎛⎝x 3⎞⎠. This chain reaction gives us hints as to what is involved in computing the derivative of sin⎛⎝x 3⎞⎠. First of all, a
change in x forcing a change in x 3 suggests that somehow the derivative of x 3 is involved. In addition, the change in x 3
forcing a change in sin⎛⎝x 3⎞⎠ suggests that the derivative of sin(u) with respect to u, where u = x 3, is also part of the
final derivative.
We can take a more formal look at the derivative of h(x) = sin ⎛⎝x 3⎞⎠ by setting up the limit that would give us the derivative
at a specific value a in the domain of h(x) = sin ⎛⎝x 3⎞⎠.
h′ (a) = xlim
→a
sin ⎛⎝x 3⎞⎠ − sin⎛⎝a 3⎞⎠
.
x−a
This expression does not seem particularly helpful; however, we can modify it by multiplying and dividing by the
expression x 3 − a 3 to obtain
h′ (a) = xlim
→a
sin ⎛⎝x 3⎞⎠ − sin⎛⎝a 3⎞⎠ x 3 − a 3
· x−a .
x3 − a3
From the definition of the derivative, we can see that the second factor is the derivative of x 3 at x = a. That is,
3
2
d ⎛ 3⎞
a3
lim xx −
− a = dx ⎝x ⎠ = 3a .
x→a
However, it might be a little more challenging to recognize that the first term is also a derivative. We can see this by letting
u = x 3 and observing that as x → a, u → a 3 :
288
Chapter 3 | Derivatives
lim
x→a
sin ⎛⎝x 3⎞⎠ − sin ⎛⎝a 3⎞⎠
x3 − a3
= lim
sinu − sin ⎛⎝a 3⎞⎠
u → a3
u − a3
= d (sinu) u = a 3
du
= cos ⎛⎝a 3).
Thus, h′ (a) = cos⎛⎝a 3⎞⎠ · 3a 2.
In other words, if h(x) = sin ⎛⎝x 3⎞⎠, then h′ (x) = cos ⎛⎝x 3⎞⎠ · 3x 2. Thus, if we think of h(x) = sin⎛⎝x 3⎞⎠ as the composition
⎛
⎝
f ∘ g⎞⎠(x) = f ⎛⎝g(x)⎞⎠ where f (x) = sin x and g(x) = x 3, then the derivative of h(x) = sin⎛⎝x 3⎞⎠ is the product of the
derivative of g(x) = x 3 and the derivative of the function f (x) = sin x evaluated at the function g(x) = x 3. At this point,
we anticipate that for h(x) = sin ⎛⎝g(x)⎞⎠, it is quite likely that h′(x) = cos(g(x))g′(x). As we determined above, this is the
case for h(x) = sin ⎛⎝x 3⎞⎠.
Now that we have derived a special case of the chain rule, we state the general case and then apply it in a general form to
other composite functions. An informal proof is provided at the end of the section.
Rule: The Chain Rule
Let f and g be functions. For all x in the domain of g for which g is differentiable at x and f is differentiable at
g(x), the derivative of the composite function
h(x) = ⎛⎝ f ∘ g⎞⎠(x) = f ⎛⎝g(x)⎞⎠
is given by
h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x).
(3.17)
Alternatively, if y is a function of u, and u is a function of x, then
dy dy du
=
· .
dx du dx
Watch an animation (http://www.openstaxcollege.org/l/20_chainrule2) of the chain rule.
Problem-Solving Strategy: Applying the Chain Rule
1. To differentiate h(x) = f ⎛⎝g(x)⎞⎠, begin by identifying f (x) and g(x).
2. Find f ′(x) and evaluate it at g(x) to obtain f ′ ⎛⎝g(x)⎞⎠.
3. Find g′(x).
4. Write h′ (x) = f ′ ⎛⎝g(x)⎞⎠ · g′ (x).
Note: When applying the chain rule to the composition of two or more functions, keep in mind that we work our way
from the outside function in. It is also useful to remember that the derivative of the composition of two functions can
be thought of as having two parts; the derivative of the composition of three functions has three parts; and so on. Also,
remember that we never evaluate a derivative at a derivative.
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Chapter 3 | Derivatives
289
The Chain and Power Rules Combined
We can now apply the chain rule to composite functions, but note that we often need to use it with other rules. For example,
to find derivatives of functions of the form h(x) = (g(x)) n, we need to use the chain rule combined with the power rule. To
do so, we can think of h(x) = ⎛⎝g(x)⎞⎠ n as f ⎛⎝g(x)⎞⎠ where f (x) = x n. Then f ′ (x) = nx n − 1. Thus, f ′ ⎛⎝g(x)⎞⎠ = n⎛⎝g(x)⎞⎠ n − 1.
This leads us to the derivative of a power function using the chain rule,
h′ (x) = n⎛⎝g(x)⎞⎠ n − 1 g′ (x)
Rule: Power Rule for Composition of Functions
For all values of x for which the derivative is defined, if
h(x) = ⎛⎝g(x)⎞⎠ n.
Then
h′ (x) = n⎛⎝g(x)⎞⎠ n − 1 g′ (x).
Example 3.48
Using the Chain and Power Rules
Find the derivative of h(x) =
⎛ 2
⎝3x
1
+ 1⎞⎠
2
.
Solution
First, rewrite h(x) =
⎛ 2
⎝3x
1
−2
2
⎞
+ 1⎠
= ⎛⎝3x 2 + 1⎞⎠ .
Applying the power rule with g(x) = 3x 2 + 1, we have
h′ (x) = −2⎛⎝3x 2 + 1⎞⎠
−3
(6x).
Rewriting back to the original form gives us
h′ (x) =
3.34
−12x .
(3x 2 + 1) 3
4
Find the derivative of h(x) = ⎛⎝2x 3 + 2x − 1⎞⎠ .
Example 3.49
Using the Chain and Power Rules with a Trigonometric Function
Find the derivative of h(x) = sin 3 x.
(3.18)
290
Chapter 3 | Derivatives
Solution
First recall that sin 3 x = (sin x) 3, so we can rewrite h(x) = sin 3 x as h(x) = (sin x) 3.
Applying the power rule with g(x) = sin x, we obtain
h′ (x) = 3(sin x) 2 cos x = 3sin 2 x cos x.
Example 3.50
Finding the Equation of a Tangent Line
Find the equation of a line tangent to the graph of h(x) =
1
at x = 2.
(3x − 5) 2
Solution
Because we are finding an equation of a line, we need a point. The x-coordinate of the point is 2. To find the
y-coordinate, substitute 2 into h(x). Since h(2) =
1
= 1, the point is (2, 1).
3(2) − 5⎞⎠ 2
⎛
⎝
For the slope, we need h′(2). To find h′(x), first we rewrite h(x) = (3x − 5) −2 and apply the power rule to
obtain
h′ (x) = −2(3x − 5) −3 (3) = −6(3x − 5) −3.
By substituting, we have h′ (2) = −6⎛⎝3(2) − 5⎞⎠ −3 = −6. Therefore, the line has equation y − 1 = −6(x − 2).
Rewriting, the equation of the line is y = −6x + 13.
3.35
3
Find the equation of the line tangent to the graph of f (x) = ⎛⎝x 2 − 2⎞⎠ at x = −2.
Combining the Chain Rule with Other Rules
Now that we can combine the chain rule and the power rule, we examine how to combine the chain rule with the other rules
we have learned. In particular, we can use it with the formulas for the derivatives of trigonometric functions or with the
product rule.
Example 3.51
Using the Chain Rule on a General Cosine Function
Find the derivative of h(x) = cos ⎛⎝g(x)⎞⎠.
Solution
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Chapter 3 | Derivatives
Think
⎛
⎝
of
291
h(x) = cos(g(x))
⎛
⎝
⎞
⎠
as
f ⎛⎝g(x)⎞⎠
where
⎞
⎠
f (x) = cos x.
Since
f ′ (x) = −sin x.
f ′ g(x) = −sin g(x) . Then we do the following calculation.
h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x)
= −sin ⎛⎝g(x)⎞⎠g′ (x)
Apply the chain rule.
Substitute f ′ ⎛⎝g(x)⎞⎠ = −sin ⎛⎝g(x)⎞⎠.
Thus, the derivative of h(x) = cos ⎛⎝g(x)⎞⎠ is given by h′ (x) = −sin ⎛⎝g(x)⎞⎠g′ (x).
In the following example we apply the rule that we have just derived.
Example 3.52
Using the Chain Rule on a Cosine Function
Find the derivative of h(x) = cos ⎛⎝5x 2⎞⎠.
Solution
Let g(x) = 5x 2. Then g′ (x) = 10x. Using the result from the previous example,
h′ (x) = −sin ⎛⎝5x 2⎞⎠ · 10x
= −10x sin ⎛⎝5x 2⎞⎠.
Example 3.53
Using the Chain Rule on Another Trigonometric Function
⎛
⎞
Find the derivative of h(x) = sec ⎝4x 5 + 2x⎠.
Solution
Apply the chain rule to h(x) = sec ⎛⎝g(x)⎞⎠ to obtain
h′ (x) = sec(g(x)tan ⎛⎝g(x)⎞⎠g′ (x).
In this problem, g(x) = 4x 5 + 2x, so we have g′ (x) = 20x 4 + 2. Therefore, we obtain
h′ (x) = sec ⎛⎝4x 5 + 2x⎞⎠ tan ⎛⎝4x 5 + 2x⎞⎠⎛⎝20x 4 + 2⎞⎠
= (20x 4 + 2)sec ⎛⎝4x 5 + 2x⎞⎠ tan ⎛⎝4x 5 + 2x⎞⎠.
3.36
Find the derivative of h(x) = sin(7x + 2).
we
have
292
Chapter 3 | Derivatives
At this point we provide a list of derivative formulas that may be obtained by applying the chain rule in conjunction
with the formulas for derivatives of trigonometric functions. Their derivations are similar to those used in Example 3.51
and Example 3.53. For convenience, formulas are also given in Leibniz’s notation, which some students find easier to
remember. (We discuss the chain rule using Leibniz’s notation at the end of this section.) It is not absolutely necessary to
memorize these as separate formulas as they are all applications of the chain rule to previously learned formulas.
Theorem 3.10: Using the Chain Rule with Trigonometric Functions
For all values of x for which the derivative is defined,
d sinu
dx
d cosu
dx
d tanu
dx
d cot u
dx
d secu
dx
d cscu
dx
d ⎛⎝sin(g(x)⎞⎠ = cos ⎛⎝g(x)⎞⎠g′(x)
dx
d ⎛⎝cos(g(x)⎞⎠ = −sin ⎛⎝g(x)⎞⎠g′(x)
dx
d ⎛⎝tan(g(x)⎞⎠ = sec 2 ⎛⎝g(x)⎞⎠g′(x)
dx
d ⎛⎝cot(g(x)⎞⎠ = −csc 2 ⎛⎝g(x)⎞⎠g′(x)
dx
d ⎛⎝sec(g(x)⎞⎠ = sec(g(x)tan ⎛⎝g(x)⎞⎠g′(x)
dx
d ⎛⎝csc(g(x)⎞⎠ = −csc(g(x))cot ⎛⎝g(x)⎞⎠g′(x)
dx
= cosu du
dx
= −sinu du
dx
2 du
= sec u
dx
2 du
= −csc u
dx
= secutanu du
dx
= −cscucot u du .
dx
Example 3.54
Combining the Chain Rule with the Product Rule
Find the derivative of h(x) = (2x + 1) 5 (3x − 2) 7.
Solution
First apply the product rule, then apply the chain rule to each term of the product.
h′ (x) = d ⎛⎝(2x + 1) 5⎞⎠ · (3x − 2) 7 + d ⎛⎝(3x − 2) 7⎞⎠ · (2x + 1) 5
dx
dx
= 5(2x + 1) 4 · 2 · (3x − 2) 7 + 7(3x − 2) 6 · 3 · (2x + 1) 5
4
6
7
= 10(2x + 1) (3x − 2) + 21(3x − 2) (2x + 1)
5
= (2x + 1) 4 (3x − 2) 6 (10(3x − 7) + 21(2x + 1))
= (2x + 1) 4 (3x − 2) 6(72x − 49)
3.37
Find the derivative of h(x) =
Apply the product rule.
Apply the chain rule.
Simplify.
Factor out (2x + 1) 4 (3x − 2) 6.
Simplify.
x
.
(2x + 3) 3
Composites of Three or More Functions
We can now combine the chain rule with other rules for differentiating functions, but when we are differentiating the
composition of three or more functions, we need to apply the chain rule more than once. If we look at this situation in
general terms, we can generate a formula, but we do not need to remember it, as we can simply apply the chain rule multiple
times.
In general terms, first we let
k(x) = h⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠.
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Chapter 3 | Derivatives
293
Then, applying the chain rule once we obtain
k′ (x) = d ⎛⎝h( f ⎛⎝g(x)⎞⎠⎞⎠ = h′⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠ · d f ⎛⎝⎛⎝g(x)⎞⎠⎞⎠.
dx
dx
Applying the chain rule again, we obtain
k′ (x) = h′ ⎛⎝ f ⎛⎝g(x)⎞⎠ f ′ ⎛⎝g(x)⎞⎠g′ (x)⎞⎠.
Rule: Chain Rule for a Composition of Three Functions
For all values of x for which the function is differentiable, if
k(x) = h⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠,
then
k′ (x) = h′ ⎛⎝ f ⎛⎝g(x)⎞⎠⎞⎠ f ′ ⎛⎝g(x)⎞⎠g′ (x).
In other words, we are applying the chain rule twice.
Notice that the derivative of the composition of three functions has three parts. (Similarly, the derivative of the composition
of four functions has four parts, and so on.) Also, remember, we can always work from the outside in, taking one derivative
at a time.
Example 3.55
Differentiating a Composite of Three Functions
Find the derivative of k(x) = cos 4 ⎛⎝7x 2 + 1⎞⎠.
Solution
First, rewrite k(x) as
4
k(x) = ⎛⎝cos ⎛⎝7x 2 + 1⎞⎠⎞⎠ .
Then apply the chain rule several times.
3⎛
⎞
k′ (x) = 4⎛⎝cos ⎛⎝7x 2 + 1⎞⎠⎞⎠ ⎝ d (cos ⎛⎝7x 2 + 1⎞⎠⎠
dx
⎛
3
Apply the chain rule.
= 4⎛⎝cos ⎛⎝7x 2 + 1⎞⎠⎞⎠ ⎛⎝−sin ⎛⎝7x 2 + 1⎞⎠⎞⎠(14x)
Apply the chain rule.
2
3
2
= −56x sin(7x + 1)cos (7x + 1)
Find the derivative of h(x) = sin 6⎛⎝x 3⎞⎠.
Example 3.56
⎞
= 4⎛⎝cos ⎛⎝7x 2 + 1⎞⎠⎞⎠ ⎛⎝−sin ⎛⎝7x 2 + 1⎞⎠⎞⎠⎝ d ⎛⎝7x 2 + 1⎞⎠⎠
dx
3
3.38
Apply the chain rule.
Simplify.
294
Chapter 3 | Derivatives
Using the Chain Rule in a Velocity Problem
A particle moves along a coordinate axis. Its position at time t is given by s(t) = sin(2t) + cos(3t). What is the
velocity of the particle at time t = π ?
6
Solution
To find v(t), the velocity of the particle at time t, we must differentiate s(t). Thus,
v(t) = s′ (t) = 2cos(2t) − 3sin(3t).
⎛ ⎞
Substituting t = π into v(t), we obtain v⎝π ⎠ = −2.
6
6
3.39
A particle moves along a coordinate axis. Its position at time t is given by s(t) = sin(4t). Find its
acceleration at time t.
Proof
At this point, we present a very informal proof of the chain rule. For simplicity’s sake we ignore certain issues: For example,
we assume that g(x) ≠ g(a) for x ≠ a in some open interval containing a. We begin by applying the limit definition of
the derivative to the function h(x) to obtain h′(a):
h′ (a) = xlim
→a
f ⎛⎝g(x)⎞⎠ − f ⎛⎝g(a)⎞⎠
.
x−a
Rewriting, we obtain
h′ (a) = xlim
→a
f ⎛⎝g(x)⎞⎠ − f ⎛⎝g(a)⎞⎠ g(x) − g(a)
· x−a .
g(x) − g(a)
Although it is clear that
lim
x→a
g(x) − g(a)
= g′(a),
x−a
it is not obvious that
lim
x→a
f ⎛⎝g(x)⎞⎠ − f ⎛⎝g(a)⎞⎠
= f ′ ⎛⎝g(a)⎞⎠.
g(x) − g(a)
To see that this is true, first recall that since g is differentiable at a, g is also continuous at a. Thus,
lim g(x) = g(a).
x→a
Next, make the substitution y = g(x) and b = g(a) and use change of variables in the limit to obtain
lim
x→a
f ⎛⎝g(x)⎞⎠ − f ⎛⎝g(a)⎞⎠
f (y) − f (b)
= lim
= f ′ (b) = f ′ ⎛⎝g(a)⎞⎠.
g(x) − g(a)
y−b
y→b
Finally,
h′ (a) = xlim
→a
f ⎛⎝g(x)⎞⎠ − f ⎛⎝g(a)⎞⎠ g(x) − g(a)
· x−a
= f ′ ⎛⎝g(a)⎞⎠g′ (a).
g(x) − g(a)
□
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Chapter 3 | Derivatives
295
Example 3.57
Using the Chain Rule with Functional Values
Let h(x) = f ⎛⎝g(x)⎞⎠. If g(1) = 4, g′ (1) = 3, and f ′ (4) = 7, find h′ (1).
Solution
Use the chain rule, then substitute.
h′ (1) = f ′ ⎛⎝g(1)⎞⎠g′ (1)
= f ′ (4) · 3
= 7·3
= 21
3.40
Apply the chain rule.
Substitute g(1) = 4 and g′ (1) = 3.
Substitute f ′(4) = 7.
Simplify.
Given h(x) = f ⎛⎝g(x)⎞⎠. If g(2) = −3, g′ (2) = 4, and f ′ (−3) = 7, find h′ (2).
The Chain Rule Using Leibniz’s Notation
As with other derivatives that we have seen, we can express the chain rule using Leibniz’s notation. This notation for the
chain rule is used heavily in physics applications.
For h(x) = f ⎛⎝g(x)⎞⎠, let u = g(x) and y = h(x) = g(u). Thus,
h′ (x) =
dy
dy
, f ′ ⎛⎝g(x)⎞⎠ = f ′ (u) =
and g′ (x) = du .
dx
dx
du
Consequently,
dy
dy du
= h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x) =
· .
dx
du dx
Rule: Chain Rule Using Leibniz’s Notation
If y is a function of u, and u is a function of x, then
dy dy du
=
· .
dx du dx
Example 3.58
Taking a Derivative Using Leibniz’s Notation, Example 1
⎛
Find the derivative of y = ⎝
x ⎞ .
3x + 2 ⎠
5
Solution
First, let u =
x . Thus, y = u 5. Next, find du and dy . Using the quotient rule,
3x + 2
dx
du
296
Chapter 3 | Derivatives
du =
2
dx (3x + 2) 2
and
dy
= 5u 4.
du
Finally, we put it all together.
dy
dy du
=
·
dx
du dx
= 5u 4 ·
⎛
= 5⎝
=
Apply the chain rule.
2
(3x + 2) 2
Substitute
x ⎞ ·
2
3x + 2 ⎠ (3x + 2) 2
4
10x 4
(3x + 2) 6
dy
2
= 5u 4 and du =
.
du
dx (3x + 2) 2
Substitute u =
x .
3x + 2
Simplify.
It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be
expressed entirely in terms of the original variable given in the problem.
Example 3.59
Taking a Derivative Using Leibniz’s Notation, Example 2
Find the derivative of y = tan ⎛⎝4x 2 − 3x + 1⎞⎠.
Solution
dy
First, let u = 4x 2 − 3x + 1. Then y = tanu. Next, find du and
:
dx
du
du = 8x − 3 and dy = sec 2 u.
dx
du
Finally, we put it all together.
dy
dy du
=
·
dx
du dx
= sec 2 u · (8x − 3)
= sec 2(4x 2 − 3x + 1) · (8x − 3)
3.41
Apply the chain rule.
dy
Use du = 8x − 3 and
= sec 2 u.
dx
du
Substitute u = 4x 2 − 3x + 1.
Use Leibniz’s notation to find the derivative of y = cos ⎛⎝x 3⎞⎠. Make sure that the final answer is
expressed entirely in terms of the variable x.
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Chapter 3 | Derivatives
297
3.6 EXERCISES
For the following exercises, given
y = f (u)
and
dy
by using Leibniz’s notation for the
dx
dy dy du
chain rule:
=
.
dx du dx
u = g(x), find
230. y = cos 3 (πx)
231. y = ⎛⎝2x 3 − x 2 + 6x + 1⎞⎠
232. y =
214. y = 3u − 6, u = 2x 2
3
1
sin 2(x)
215. y = 6u 3, u = 7x − 4
233. y = (tan x + sin x) −3
216. y = sinu, u = 5x − 1
234. y = x 2 cos 4 x
217. y = cosu, u = −x
235. y = sin(cos7x)
218. y = tanu, u = 9x + 2
236. y = 6 + secπx 2
219. y = 4u + 3, u = x 2 − 6x
237. y = cot 3 (4x + 1)
For each of the following exercises,
238. Let y = ⎡⎣ f (x)⎤⎦ 3 and suppose that f ′ (1) = 4 and
8
a. decompose each function in the form y = f (u)
and u = g(x), and
b. find
dy
as a function of x.
dx
220. y = (3x − 2)
3
241.
4
and
suppose
that
dy
= 3 when x = −1. Find f ′ (−1)
dx
y = ⎛⎝ f (u) + 3x⎞⎠ 2
Let
f (4) = 6 and
and
u = x 3 − 2x.
If
dy
= 18 when x = 2, find f ′ (4).
dx
[T] Find the equation of the tangent line to
⎛ ⎞
y = −sin ⎝ x ⎠ at the origin. Use a calculator to graph the
2
7
function and the tangent line together.
224. y = tan(sec x)
242.
[T] Find the equation of the tangent line to
2
y = ⎛⎝3x + 1x ⎞⎠ at the point (1, 16). Use a calculator to
225. y = csc(πx + 1)
graph the function and the tangent line together.
226. y = cot 2 x
227. y = −6sin
y = ⎛⎝ f (x) + 5x 2⎞⎠
Let
240.
222. y = sin 5 (x)
⎞
⎛
223. y = ⎝ x + 7x ⎠
7
239.
f (−1) = −4 and
6
221. y = ⎛⎝3x 2 + 1⎞⎠
dy
= 10 for x = 1. Find f (1).
dx
243. Find the x -coordinates at which the tangent line to
−3
⎛
x
⎞
8
y = ⎝x − 6x ⎠ is horizontal.
For the following exercises, find
4
⎞
dy
for each function.
dx
244. [T] Find an equation of the line that is normal to
⎛
⎞
g(θ) = sin 2 (πθ) at the point ⎝1 , 1 ⎠. Use a calculator to
4 2
228. y = ⎛⎝3x 2 + 3x − 1⎠
graph the function and the normal line together.
229. y = (5 − 2x) −2
For the following exercises, use the information in the
following table to find h′(a) at the given value for a.
298
Chapter 3 | Derivatives
x
f(x)
f′(x)
g(x)
g′(x)
0
2
5
0
2
1
1
−2
3
0
2
4
4
1
−1
3
3
−3
2
3
255. [T] The total cost to produce x boxes of Thin Mint
Girl
Scout
cookies
is
C
dollars,
where
C = 0.0001x 3 − 0.02x 2 + 3x + 300. In t weeks
production is estimated to be x = 1600 + 100t boxes.
a. Find the marginal cost C′ (x).
b. Use Leibniz’s notation for the chain rule,
dC = dC · dx , to find the rate with respect to
dt
dx dt
time t that the cost is changing.
c. Use b. to determine how fast costs are increasing
when t = 2 weeks. Include units with the answer.
256. [T] The formula for the area of a circle is A = πr 2,
245. h(x) = f ⎛⎝g(x)⎞⎠; a = 0
246. h(x) = g⎛⎝ f (x)⎞⎠; a = 0
−2
⎞
247. h(x) = ⎛⎝x 4 + g(x)⎠
⎛ f (x) ⎞
;a=3
g(x) ⎠
249. h(x) = f ⎛⎝x + f (x)⎞⎠; a = 1
250. h(x) = ⎛⎝1 + g(x)⎞⎠ 3; a = 2
251. h(x) =
g⎛⎝2
+
f ⎛⎝x 2⎞⎠⎞⎠;
a. Suppose r = 2 −
;a=1
2
248. h(x) = ⎝
where r is the radius of the circle. Suppose a circle is
expanding, meaning that both the area A and the radius r
(in inches) are expanding.
100
where t is time in
(t + 7) 2
seconds. Use the chain rule dA = dA · dr to find
dt
dr dt
the rate at which the area is expanding.
b. Use a. to find the rate at which the area is
expanding at t = 4 s.
257.
[T] The formula for the volume of a sphere is
S = 4 πr 3, where r (in feet) is the radius of the sphere.
3
Suppose a spherical snowball is melting in the sun.
a=1
252. h(x) = f ⎛⎝g(sin x)⎞⎠; a = 0
253. [T] The position function of a freight train is given by
s(t) = 100(t + 1) −2, with s in meters and t in seconds.
At time t = 6 s, find the train’s
a. velocity and
b. acceleration.
c. Using a. and b. is the train speeding up or slowing
down?
254. [T] A mass hanging from a vertical spring is in
simple harmonic motion as given by the following position
function, where t is measured in seconds and s is in
⎛
⎞
inches: s(t) = −3cos ⎝πt + π ⎠.
4
a. Determine the position of the spring at t = 1.5 s.
b. Find the velocity of the spring at t = 1.5 s.
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a. Suppose r =
1
− 1 where t is time in
(t + 1) 2 12
minutes. Use the chain rule dS = dS · dr to find
dt
dr dt
the rate at which the snowball is melting.
b. Use a. to find the rate at which the volume is
changing at t = 1 min.
258. [T] The daily temperature in degrees Fahrenheit of
Phoenix in the summer can be modeled by the function
⎡
⎤
T(x) = 94 − 10cos⎣ π (x − 2)⎦, where x is hours after
12
midnight. Find the rate at which the temperature is
changing at 4 p.m.
259. [T] The depth (in feet) of water at a dock changes
with the rise and fall of tides. The depth is modeled by
⎛
⎞
the function D(t) = 5sin ⎝π t − 7π ⎠ + 8, where t is the
6
6
number of hours after midnight. Find the rate at which the
depth is changing at 6 a.m.
Chapter 3 | Derivatives
299
3.7 | Derivatives of Inverse Functions
Learning Objectives
3.7.1 Calculate the derivative of an inverse function.
3.7.2 Recognize the derivatives of the standard inverse trigonometric functions.
In this section we explore the relationship between the derivative of a function and the derivative of its inverse. For functions
whose derivatives we already know, we can use this relationship to find derivatives of inverses without having to use the
limit definition of the derivative. In particular, we will apply the formula for derivatives of inverse functions to trigonometric
functions. This formula may also be used to extend the power rule to rational exponents.
The Derivative of an Inverse Function
We begin by considering a function and its inverse. If f (x) is both invertible and differentiable, it seems reasonable that
the inverse of f (x) is also differentiable. Figure 3.28 shows the relationship between a function f (x) and its inverse
p
f −1 (x). Look at the point ⎛⎝a, f −1 (a)⎞⎠ on the graph of f −1(x) having a tangent line with a slope of ⎛⎝ f −1⎞⎠′ (a) = q . This
q
point corresponds to a point ⎛⎝ f −1 (a), a⎞⎠ on the graph of f (x) having a tangent line with a slope of f ′ ⎛⎝ f −1 (a)⎞⎠ = p .
Thus, if f −1(x) is differentiable at a, then it must be the case that
⎛ −1⎞
⎝f
⎠′ (a)
=
1
.
f ′ ⎛⎝ f −1 (a)⎞⎠
Figure 3.28 The tangent lines of a function and its inverse are
related; so, too, are the derivatives of these functions.
We may also derive the formula for the derivative of the inverse by first recalling that x = f ⎛⎝ f −1 (x)⎞⎠. Then by
differentiating both sides of this equation (using the chain rule on the right), we obtain
1 = f ′ ⎛⎝ f −1 (x)⎞⎠⎛⎝ f −1 )′(x)⎞⎠.
Solving for ( f −1 )′(x), we obtain
⎛ −1⎞
⎝f
⎠′ (x)
We summarize this result in the following theorem.
=
1
.
f ′ ⎛⎝ f −1 (x)⎞⎠
(3.19)
300
Chapter 3 | Derivatives
Theorem 3.11: Inverse Function Theorem
Let f (x) be a function that is both invertible and differentiable. Let y = f −1 (x) be the inverse of f (x). For all x
satisfying f ′ ⎛⎝ f −1 (x)⎞⎠ ≠ 0,
dy
1
= d ⎛ f −1(x)⎞⎠ = ⎛⎝ f −1⎞⎠′ (x) = ⎛ −1
.
dx dx ⎝
f ′ ⎝ f (x)⎞⎠
Alternatively, if y = g(x) is the inverse of f (x), then
g(x) =
1 .
f ′ ⎛⎝g(x)⎞⎠
Example 3.60
Applying the Inverse Function Theorem
2
Use the inverse function theorem to find the derivative of g(x) = x +
x . Compare the resulting derivative to that
obtained by differentiating the function directly.
Solution
2
2
1
The inverse of g(x) = x +
x is f (x) = x − 1 . Since g′ (x) = f ′ ⎛⎝g(x)⎞⎠ , begin by finding f ′ (x). Thus,
f ′ (x) =
−2 and f ′ ⎛⎝g(x)⎞⎠ =
−2
=
⎛
⎞2
(x − 1) 2
⎝g(x) − 1⎠
−2
⎛x + 2
⎝ x
2
⎞
− 1⎠
2
= −x .
2
Finally,
g′ (x) =
1
= − 22 .
f ′ ⎛⎝g(x)⎞⎠
x
We can verify that this is the correct derivative by applying the quotient rule to g(x) to obtain
g′ (x) = − 22 .
x
3.42
Use the inverse function theorem to find the derivative of g(x) =
by differentiating g(x) directly.
Example 3.61
Applying the Inverse Function Theorem
Use the inverse function theorem to find the derivative of g(x) = 3 x.
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1 . Compare the result obtained
x+2
Chapter 3 | Derivatives
301
Solution
1 , begin by finding
f ′ ⎛⎝g(x)⎞⎠
The function g(x) = 3 x is the inverse of the function f (x) = x 3. Since g′ (x) =
f ′ (x). Thus,
2
f ′ (x) = 3x 3 and f ′ ⎛⎝g(x)⎞⎠ = 3⎛⎝3 x⎞⎠ = 3x 2/3.
Finally,
g′ (x) =
3.43
1 = 1 x −2/3.
3x 2/3 3
Find the derivative of g(x) = 5 x by applying the inverse function theorem.
From the previous example, we see that we can use the inverse function theorem to extend the power rule to exponents of
q
the form 1
n , where n is a positive integer. This extension will ultimately allow us to differentiate x , where q is any
rational number.
Theorem 3.12: Extending the Power Rule to Rational Exponents
The power rule may be extended to rational exponents. That is, if n is a positive integer, then
d ⎛x 1/n⎞ = 1 x (1/n) − 1.
⎠
n
dx ⎝
(3.20)
Also, if n is a positive integer and m is an arbitrary integer, then
d ⎛x m/n⎞ = m x (m/n) − 1.
⎠
n
dx ⎝
(3.21)
Proof
The function g(x) = x 1/n is the inverse of the function f (x) = x n. Since g′ (x) =
1 , begin by finding f ′ (x).
f ′ ⎛⎝g(x)⎞⎠
Thus,
f ′ (x) = nx n − 1 and f ′ ⎛⎝g(x)⎞⎠ = n(x 1/n) n − 1 = nx
(n − 1)/n
.
Finally,
g′ (x) =
nx
1
(n − 1)/n
To differentiate x m/n we must rewrite it as ⎛⎝x 1/n⎞⎠
m
(1 − n)/n
(1/n) − 1
= 1n x
= 1n x
.
and apply the chain rule. Thus,
m−1
(1/n) − 1
(m/n) − 1
d ⎛x m/n⎞ = d ⎛⎛x 1/n⎞m⎞ = m⎛x 1/n⎞
· 1n x
=m
.
⎠
⎠
⎝
⎠
⎝
⎝
nx
⎠
dx ⎝
dx
□
302
Chapter 3 | Derivatives
Example 3.62
Applying the Power Rule to a Rational Power
Find the equation of the line tangent to the graph of y = x 2/3 at x = 8.
Solution
dy
and evaluate it at x = 8. Since
dx
First find
dy 2 −1/3
dy
= x
and
dx 3
dx
the slope of the tangent line to the graph at x = 8 is 1 .
|
x=8
=1
3
3
Substituting x = 8 into the original function, we obtain y = 4. Thus, the tangent line passes through the point
(8, 4). Substituting into the point-slope formula for a line, we obtain the tangent line
y = 1 x + 4.
3
3
3.44
Find the derivative of s(t) = 2t + 1.
Derivatives of Inverse Trigonometric Functions
We now turn our attention to finding derivatives of inverse trigonometric functions. These derivatives will prove invaluable
in the study of integration later in this text. The derivatives of inverse trigonometric functions are quite surprising in that
their derivatives are actually algebraic functions. Previously, derivatives of algebraic functions have proven to be algebraic
functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Here, for the first time,
we see that the derivative of a function need not be of the same type as the original function.
Example 3.63
Derivative of the Inverse Sine Function
Use the inverse function theorem to find the derivative of g(x) = sin −1 x.
Solution
⎡
⎤
Since for x in the interval ⎣− π , π ⎦, f (x) = sin x is the inverse of g(x) = sin −1 x, begin by finding f ′(x).
2 2
Since
f ′ (x) = cos x and f ′ ⎛⎝g(x)⎞⎠ = cos ⎛⎝sin −1 x⎞⎠ = 1 − x 2,
we see that
1 .
g′ (x) = d ⎛⎝sin −1 x⎞⎠ = ⎛ 1 ⎞ =
dx
f ′ ⎝g(x)⎠
1 − x2
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Chapter 3 | Derivatives
303
Analysis
To see that cos ⎛⎝sin −1 x⎞⎠ = 1 − x 2, consider the following argument. Set sin −1 x = θ. In this case, sinθ = x
where − π ≤ θ ≤ π . We begin by considering the case where 0 < θ < π . Since θ is an acute angle, we may
2
2
2
construct a right triangle having acute angle θ, a hypotenuse of length 1 and the side opposite angle θ having
length x. From the Pythagorean theorem, the side adjacent to angle θ has length
1 − x 2. This triangle is
shown in Figure 3.29. Using the triangle, we see that cos ⎛⎝sin −1 x⎞⎠ = cosθ = 1 − x 2.
Figure 3.29 Using a right triangle having acute angle θ, a
hypotenuse of length 1, and the side opposite angle θ having
⎛
⎞
length x, we can see that cos ⎝sin −1 x⎠ = cosθ = 1 − x 2.
In the case where − π < θ < 0, we make the observation that 0 < −θ < π and hence
2
2
cos ⎛⎝sin −1 x⎞⎠ = cosθ = cos(−θ) = 1 − x 2.
Now if θ = π or θ = − π , x = 1 or x = −1, and since in either case cosθ = 0 and 1 − x 2 = 0, we have
2
2
cos ⎛⎝sin −1 x⎞⎠ = cosθ = 1 − x 2.
Consequently, in all cases, cos ⎛⎝sin −1 x⎞⎠ = 1 − x 2.
Example 3.64
Applying the Chain Rule to the Inverse Sine Function
Apply the chain rule to the formula derived in Example 3.61 to find the derivative of h(x) = sin −1 ⎛⎝g(x)⎞⎠ and
use this result to find the derivative of h(x) = sin −1 ⎛⎝2x 3⎞⎠.
Solution
Applying the chain rule to h(x) = sin −1 ⎛⎝g(x)⎞⎠, we have
h′ (x) =
1
g′ (x).
1 − ⎛⎝g(x)⎞⎠ 2
Now let g(x) = 2x 3, so g′ (x) = 6x. Substituting into the previous result, we obtain
304
Chapter 3 | Derivatives
1
· 6x
1 − 4x 6
6x .
=
1 − 4x 6
h′ (x) =
3.45
Use the inverse function theorem to find the derivative of g(x) = tan −1 x.
The derivatives of the remaining inverse trigonometric functions may also be found by using the inverse function theorem.
These formulas are provided in the following theorem.
Theorem 3.13: Derivatives of Inverse Trigonometric Functions
d sin −1 x =
1
dx
1 − (x) 2
d cos −1 x =
−1
dx
1 − (x) 2
d tan −1 x =
1
dx
1 + (x) 2
d cot −1 x = −1
dx
1 + (x) 2
d sec −1 x =
1
dx
|x| (x) 2 − 1
d csc −1 x =
−1
dx
|x| (x) 2 − 1
Example 3.65
Applying Differentiation Formulas to an Inverse Tangent Function
Find the derivative of f (x) = tan −1 ⎛⎝x 2⎞⎠.
Solution
Let g(x) = x 2, so g′ (x) = 2x. Substituting into Equation 3.24, we obtain
f ′ (x) =
1
1 + ⎛⎝x 2⎞⎠
2
· (2x).
Simplifying, we have
f ′ (x) =
2x .
1 + x4
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(3.22)
(3.23)
(3.24)
(3.25)
(3.26)
(3.27)
Chapter 3 | Derivatives
305
Example 3.66
Applying Differentiation Formulas to an Inverse Sine Function
Find the derivative of h(x) = x 2 sin −1 x.
Solution
By applying the product rule, we have
1
· x 2.
2
1−x
h′ (x) = 2x sin −1 x +
3.46
Find the derivative of h(x) = cos −1 (3x − 1).
Example 3.67
Applying the Inverse Tangent Function
⎛ ⎞
The position of a particle at time t is given by s(t) = tan −1 ⎝1t ⎠ for t ≥ 1 . Find the velocity of the particle at
2
time t = 1.
Solution
Begin by differentiating s(t) in order to find v(t). Thus,
v(t) = s′ (t) =
1
1+
⎛1 ⎞
⎝t ⎠
2
· −1
.
t2
Simplifying, we have
v(t) = −
1 .
t2 + 1
Thus, v(1) = − 1 .
2
3.47
Find the equation of the line tangent to the graph of f (x) = sin −1 x at x = 0.
306
Chapter 3 | Derivatives
3.7 EXERCISES
For the following exercises, use the graph of y = f (x) to
263.
a. sketch the graph of y = f −1 (x), and
b. use part a. to estimate ⎛⎝ f −1⎞⎠′ (1).
260.
For the following exercises, use the functions y = f (x) to
find
261.
a.
df
at x = a and
dx
b.
x = f −1 (y).
c. Then use part b. to find
262.
d f −1
at y = f (a).
dy
264.
f (x) = 6x − 1, x = −2
265.
f (x) = 2x 3 − 3, x = 1
266.
f (x) = 9 − x 2, 0 ≤ x ≤ 3, x = 2
267.
f (x) = sin x, x = 0
For each of the following functions, find ⎛⎝ f −1⎞⎠′ (a).
268.
f (x) = x 2 + 3x + 2, x ≥ −1, a = 2
269.
f (x) = x 3 + 2x + 3, a = 0
270.
f (x) = x + x, a = 2
271.
f (x) = x − 2x , x < 0, a = 1
272.
f (x) = x + sin x, a = 0
273.
f (x) = tan x + 3x 2, a = 0
For each of the given functions y = f (x),
a. find the slope of the tangent line to its inverse
function f −1 at the indicated point P, and
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Chapter 3 | Derivatives
307
b. find the equation of the tangent line to the graph of
f −1 at the indicated point.
274.
f (x) =
4 , P(2, 1)
1 + x2
292.
f ⎛⎝ 3⎞⎠ = 1 , f ′⎛⎝ 3⎞⎠ = 2 , a = 1
2
2
3
293.
f (1) = −3, f ′(1) = 10, a = −3
294.
f (1) = 0, f ′(1) = −2, a = 0
275.
f (x) = x − 4, P(2, 8)
276.
f (x) = ⎛⎝x 3 + 1⎞⎠ , P(16, 1)
277.
f (x) = −x 3 − x + 2, P(−8, 2)
a. Find the velocity of the hockey puck at any time t.
b. Find the acceleration of the puck at any time t.
c. Evaluate a. and b. for t = 2, 4, and 6 seconds.
278.
f (x) = x 5 + 3x 3 − 4x − 8, P(−8, 1)
d. What conclusion can be drawn from the results in
c.?
295. [T] The position of a moving hockey puck after t
seconds is s(t) = tan −1 t where s is in meters.
4
For the following exercises, find
dy
for the given
dx
function.
279. y = sin −1 ⎛⎝x 2⎞⎠
296. [T] A building that is 225 feet tall casts a shadow
of various lengths x as the day goes by. An angle of
elevation θ is formed by lines from the top and bottom
of the building to the tip of the shadow, as seen in the
following figure. Find the rate of change of the angle of
elevation dθ when x = 272 feet.
dx
280. y = cos −1 ( x)
⎛ ⎞
281. y = sec −1 ⎝1x ⎠
282. y = csc −1 x
283. y = ⎛⎝1 + tan −1 x⎞⎠
3
284. y = cos −1 (2x) · sin −1 (2x)
285. y =
1
tan −1 (x)
286. y = sec −1 (−x)
287. y = cot −1 4 − x 2
288. y = x · csc −1 x
For the following exercises, use the given values to find
⎛ −1⎞
⎝f
⎠′ (a).
289.
f (π) = 0, f ′(π) = −1, a = 0
290.
f (6) = 2, f ′ (6) = 1 , a = 2
3
291.
⎛ ⎞
⎛ ⎞
f ⎝1 ⎠ = −8, f ′⎝1 ⎠ = 2, a = −8
3
3
297. [T] A pole stands 75 feet tall. An angle θ is formed
when wires of various lengths of x feet are attached from
the ground to the top of the pole, as shown in the following
figure. Find the rate of change of the angle dθ when a wire
dx
of length 90 feet is attached.
308
Chapter 3 | Derivatives
298. [T] A television camera at ground level is 2000 feet
away from the launching pad of a space rocket that is
set to take off vertically, as seen in the following figure.
The angle of elevation of the camera can be found by
⎛
⎞
θ = tan −1 ⎝ x ⎠, where x is the height of the rocket.
2000
Find the rate of change of the angle of elevation after
launch when the camera and the rocket are 5000 feet apart.
299. [T] A local movie theater with a 30-foot-high screen
that is 10 feet above a person’s eye level when seated
has a viewing angle θ (in radians) given by
θ = cot −1 x − cot −1 x , where x is the distance in
40
10
feet away from the movie screen that the person is sitting,
as shown in the following figure.
a. Find dθ .
dx
b. Evaluate dθ for x = 5, 10, 15, and 20.
dx
c. Interpret the results in b..
d. Evaluate dθ for x = 25, 30, 35, and 40
dx
e. Interpret the results in d. At what distance x should
the person stand to maximize his or her viewing
angle?
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Chapter 3 | Derivatives
309
3.8 | Implicit Differentiation
Learning Objectives
3.8.1 Find the derivative of a complicated function by using implicit differentiation.
3.8.2 Use implicit differentiation to determine the equation of a tangent line.
We have already studied how to find equations of tangent lines to functions and the rate of change of a function at a specific
point. In all these cases we had the explicit equation for the function and differentiated these functions explicitly. Suppose
instead that we want to determine the equation of a tangent line to an arbitrary curve or the rate of change of an arbitrary
curve at a point. In this section, we solve these problems by finding the derivatives of functions that define y implicitly in
terms of x.
Implicit Differentiation
In most discussions of math, if the dependent variable y is a function of the independent variable x, we express y in terms
of x. If this is the case, we say that y is an explicit function of x. For example, when we write the equation y = x 2 + 1,
we are defining y explicitly in terms of x. On the other hand, if the relationship between the function y and the variable x
is expressed by an equation where y is not expressed entirely in terms of x, we say that the equation defines y implicitly
in terms of x. For example, the equation y − x 2 = 1 defines the function y = x 2 + 1 implicitly.
Implicit differentiation allows us to find slopes of tangents to curves that are clearly not functions (they fail the vertical line
test). We are using the idea that portions of y are functions that satisfy the given equation, but that y is not actually a
function of x.
In general, an equation defines a function implicitly if the function satisfies that equation. An equation may define many
different functions implicitly. For example, the functions
⎧ 25 − x 2 if − 25 ≤ x < 0
, which are illustrated in Figure 3.30, are just three of the many
y = 25 − x 2 and y = ⎨
⎩ − 25 − x 2 if 0 ≤ x ≤ 25
functions defined implicitly by the equation x 2 + y 2 = 25.
310
Chapter 3 | Derivatives
Figure 3.30 The equation x 2 + y 2 = 25 defines many functions implicitly.
If we want to find the slope of the line tangent to the graph of x 2 + y 2 = 25 at the point (3, 4), we could evaluate
the derivative of the function y = 25 − x 2 at x = 3. On the other hand, if we want the slope of the tangent line at the
point (3, −4), we could use the derivative of y = − 25 − x 2. However, it is not always easy to solve for a function
defined implicitly by an equation. Fortunately, the technique of implicit differentiation allows us to find the derivative of
an implicitly defined function without ever solving for the function explicitly. The process of finding
differentiation is described in the following problem-solving strategy.
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dy
using implicit
dx
Chapter 3 | Derivatives
311
Problem-Solving Strategy: Implicit Differentiation
To perform implicit differentiation on an equation that defines a function y implicitly in terms of a variable x, use
the following steps:
1. Take the derivative of both sides of the equation. Keep in mind that y is a function of x. Consequently, whereas
d (sin x) = cos x, d (sin y) = cos y dy because we must use the chain rule to differentiate sin y with respect
dx
dx
dx
to x.
2. Rewrite the equation so that all terms containing
dy
dy
are on the left and all terms that do not contain
are
dx
dx
on the right.
3. Factor out
4. Solve for
dy
on the left.
dx
dy
by dividing both sides of the equation by an appropriate algebraic expression.
dx
Example 3.68
Using Implicit Differentiation
Assuming that y is defined implicitly by the equation x 2 + y 2 = 25, find
dy
.
dx
Solution
Follow the steps in the problem-solving strategy.
d ⎛x 2 + y 2⎞ = d (25)
⎠
dx ⎝
dx
d ⎛x 2⎞ + d ⎛y 2⎞ = 0
dx ⎝ ⎠ dx ⎝ ⎠
2x + 2y
2y
dy
= 0
dx
dy
= −2x
dx
dy
= − xy
dx
Step 1. Diffe entiate both sides of the equation.
Step 1.1. Use the sum rule on the left.
On the right d (25) = 0.
dx
Step 1.2. Take the derivatives, so d ⎛⎝x 2⎞⎠ = 2x
dx
dy
⎛
⎞
and d ⎝y 2⎠ = 2y .
dx
dx
dy
on the left.
Step 2. Keep the terms with
dx
Move the remaining terms to the right.
Step 4. Divide both sides of the equation by
2y. (Step 3 does not apply in this case.)
Analysis
dy
is in terms of both the independent variable x and the dependent
dx
dy
variable y. Although in some cases it may be possible to express
in terms of x only, it is generally not
dx
Note that the resulting expression for
possible to do so.
312
Chapter 3 | Derivatives
Example 3.69
Using Implicit Differentiation and the Product Rule
Assuming that y is defined implicitly by the equation x 3 sin y + y = 4x + 3, find
dy
.
dx
Solution
d ⎛x 3 sin y + y⎞ = d (4x + 3)
⎠
dx
dx ⎝
d ⎛x 3 sin y⎞ + d (y) = 4
⎠
dx
dx ⎝
⎛ d ⎛ 3⎞
3⎞ dy
⎞
d⎛
⎝dx ⎝x ⎠ · sin y + dx ⎝sin y⎠ · x ⎠ + dx = 4
dy
dy ⎞
⎛
= 4
3x 2 sin y + ⎝cos y ⎠ · x 3 +
dx
dx
x 3 cos y
dy dy
+
= 4 − 3x 2 sin y
dx dx
dy ⎛ 3
x cos y + 1⎞⎠ = 4 − 3x 2 sin y
dx ⎝
4 − 3x 2 sin y
dy
=
dx
x 3 cos y + 1
Step 1: Diffe entiate both sides of the equation.
Step 1.1: Apply the sum rule on the left.
On the right, d (4x + 3) = 4.
dx
Step 1.2: Use the product rule to fin
d ⎛x 3 sin y⎞. Observe that d (y) = dy .
⎠
dx
dx ⎝
dx
⎛ 3⎞
2
d
Step 1.3: We know ⎝x ⎠ = 3x . Use the
dx
dy
chain rule to obtain d ⎛⎝sin y⎞⎠ = cos y .
dx
dx
dy
Step 2: Keep all terms containing
on the
dx
left. Move all other terms to the right.
dy
Step 3: Factor out
on the left.
dx
dy
Step 4: Solve for
by dividing both sides of
dx
the equation by x 3 cos y + 1.
Example 3.70
Using Implicit Differentiation to Find a Second Derivative
Find
d2 y
if x 2 + y 2 = 25.
dx 2
Solution
In Example 3.68, we showed that
dy
= − xy . We can take the derivative of both sides of this equation to find
dx
d2 y
.
dx 2
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Chapter 3 | Derivatives
313
d2 y
⎛
⎞
= d ⎝− xy ⎠
dy
dx 2
= −
Diffe entiate both sides of
dy ⎞
⎛
⎝1 · y − x dx ⎠
y2
dy
= − xy .
dx
⎛
⎞
Use the quotient rule to fin d ⎝− xy ⎠.
dy
dy
=
=
=
−y + x dx
Simplify.
y2
−y + x⎛⎝− xy ⎞⎠
y2
−y 2 − x 2
y3
Substitute
dy
= − xy .
dx
Simplify.
At this point we have found an expression for
d2 y
. If we choose, we can simplify the expression further by
dx 2
recalling that x 2 + y 2 = 25 and making this substitution in the numerator to obtain
3.48
Find
d2 y
= − 253 .
dx 2
y
dy
for y defined implicitly by the equation 4x 5 + tan y = y 2 + 5x.
dx
Finding Tangent Lines Implicitly
Now that we have seen the technique of implicit differentiation, we can apply it to the problem of finding equations of
tangent lines to curves described by equations.
Example 3.71
Finding a Tangent Line to a Circle
Find the equation of the line tangent to the curve x 2 + y 2 = 25 at the point (3, −4).
Solution
Although we could find this equation without using implicit differentiation, using that method makes it much
easier. In Example 3.68, we found
dy
= − xy .
dx
The slope of the tangent line is found by substituting (3, −4) into this expression. Consequently, the slope of the
tangent line is
dy
dx
|
(3, −4)
= − 3 = 3.
−4 4
Using the point (3, −4) and the slope 3 in the point-slope equation of the line, we obtain the equation
y = 3 x − 25 (Figure 3.31).
4
4
4
314
Chapter 3 | Derivatives
Figure 3.31 The line y = 3 x − 25 is tangent to
2
2
4
4
x + y = 25 at the point (3, −4).
Example 3.72
Finding the Equation of the Tangent Line to a Curve
⎛
⎞
Find the equation of the line tangent to the graph of y 3 + x 3 − 3xy = 0 at the point ⎝3 , 3 ⎠ (Figure 3.32). This
2 2
curve is known as the folium (or leaf) of Descartes.
Figure 3.32 Finding the tangent line to the folium of
⎛
⎞
Descartes at ⎝3 , 3 ⎠.
2 2
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Chapter 3 | Derivatives
315
Solution
Begin by finding
dy
.
dx
d ⎛y 3 + x 3 − 3xy⎞ = d (0)
⎠
dx
dx ⎝
dy ⎞
2 dy
2 ⎛
3y
+ 3x − ⎝3y + 3x⎠ = 0
dx
dx
dy
3y − 3x 2
=
.
dx
3y 2 − 3x
dy 3y − 3x 2
⎛
⎞
Next, substitute ⎝3 , 3 ⎠ into
=
to find the slope of the tangent line:
2 2
dx 3y 2 − 3x
dy
dx
|
⎛3 3 ⎞
⎝2 , 2 ⎠
= −1.
Finally, substitute into the point-slope equation of the line to obtain
y = −x + 3.
Example 3.73
Applying Implicit Differentiation
In a simple video game, a rocket travels in an elliptical orbit whose path is described by the equation
4x 2 + 25y 2 = 100. The rocket can fire missiles along lines tangent to its path. The object of the game is to
destroy an incoming asteroid traveling along the positive x-axis toward (0, 0). If the rocket fires a missile when
⎛
⎞
it is located at ⎝3, 8 ⎠, where will it intersect the x-axis?
3
Solution
To solve this problem, we must determine where the line tangent to the graph of
dy
⎛
⎞
4x 2 + 25y 2 = 100 at ⎝3, 8 ⎠ intersects the x-axis. Begin by finding
implicitly.
3
dx
Differentiating, we have
8x + 50y
Solving for
dy
= 0.
dx
dy
, we have
dx
The slope of the tangent line is
dy
dx
|
dy
= − 4x .
dx
25y
⎛ 8⎞
⎝3, 3 ⎠
= − 9 . The equation of the tangent line is y = − 9 x + 183 . To
200
50
50
316
Chapter 3 | Derivatives
determine where the line intersects the x-axis, solve 0 = − 9 x + 183 . The solution is x = 61 . The missile
⎛
⎞
intersects the x-axis at the point ⎝61 , 0⎠.
3
3.49
50
200
3
Find the equation of the line tangent to the hyperbola x 2 − y 2 = 16 at the point (5, 3).
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Chapter 3 | Derivatives
317
3.8 EXERCISES
For the following exercises, use implicit differentiation to
dy
.
find
dx
316. [T] The graph of a folium of Descartes with equation
2x 3 + 2y 3 − 9xy = 0 is given in the following graph.
300. x 2 − y 2 = 4
301. 6x 2 + 3y 2 = 12
302. x 2 y = y − 7
303. 3x 3 + 9xy 2 = 5x 3
304. xy − cos(xy) = 1
a. Find the equation of the tangent line at the point
(2, 1). Graph the tangent line along with the
305. y x + 4 = xy + 8
folium.
b. Find the equation of the normal line to the tangent
line in a. at the point (2, 1).
306. −xy − 2 = x
7
307. ysin(xy) = y 2 + 2
317. For the equation x 2 + 2xy − 3y 2 = 0,
308. (xy) 2 + 3x = y 2
a. Find the equation of the normal to the tangent line
at the point (1, 1).
309. x 3 y + xy 3 = −8
b. At what other point does the normal line in a.
intersect the graph of the equation?
For the following exercises, find the equation of the tangent
line to the graph of the given equation at the indicated
point. Use a calculator or computer software to graph the
function and the tangent line.
310. [T] x 4 y − xy 3 = −2, (−1, −1)
311. [T] x 2 y 2 + 5xy = 14, (2, 1)
⎛
⎞
312. [T] tan(xy) = y, ⎝π , 1⎠
4
318. Find all points on the graph of y 3 − 27y = x 2 − 90
at which the tangent line is vertical.
319. For the equation x 2 + xy + y 2 = 7,
a. Find the x -intercept(s).
b. Find the slope of the tangent line(s) at the
x-intercept(s).
c. What does the value(s) in b. indicate about the
tangent line(s)?
320. Find the equation of the tangent line to the graph of
313. [T] xy + sin(πy) − 2x = 10, (2, −3)
⎛
⎞
the equation sin −1 x + sin −1 y = π at the point ⎝0, 1 ⎠.
2
6
314. [T] xy + 5x − 7 = − 3 y, (1, 2)
4
321. Find the equation of the tangent line to the graph of
the equation tan −1 (x + y) = x 2 + π at the point (0, 1).
⎛
⎞
315. [T] xy + sin(x) = 1, ⎝π , 0⎠
2
322. Find y′ and y″ for x 2 + 6xy − 2y 2 = 3.
2
2
4
318
Chapter 3 | Derivatives
323. [T] The number of cell phones produced when x
dollars is spent on labor and y dollars is spent on capital
invested by a manufacturer can be modeled by the equation
60x 3/4 y 1/4 = 3240.
a. Find
dy
and evaluate at the point (81, 16).
dx
b. Interpret the result of a.
324. [T] The number of cars produced when x dollars is
spent on labor and y dollars is spent on capital invested
by a manufacturer can be modeled by the equation
30x 1/3 y 2/3 = 360. (Both x and y are measured in
thousands of dollars.)
a. Find
dy
and evaluate at the point (27, 8).
dx
b. Interpret the result of a.
325. The volume of a right circular cone of radius x
and height y is given by V = 1 πx 2 y. Suppose that the
3
volume of the cone is 85π cm 3. Find
dy
when x = 4 and
dx
y = 16.
For the following exercises, consider a closed rectangular
box with a square base with side x and height y.
326. Find an equation for the surface area of the
rectangular box, S(x, y).
327. If the surface area of the rectangular box is 78 square
feet, find
dy
when x = 3 feet and y = 5 feet.
dx
For the following exercises, use implicit differentiation to
determine y′. Does the answer agree with the formulas we
have previously determined?
328. x = sin y
329. x = cos y
330. x = tan y
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Chapter 3 | Derivatives
319
3.9 | Derivatives of Exponential and Logarithmic
Functions
Learning Objectives
3.9.1 Find the derivative of exponential functions.
3.9.2 Find the derivative of logarithmic functions.
3.9.3 Use logarithmic differentiation to determine the derivative of a function.
So far, we have learned how to differentiate a variety of functions, including trigonometric, inverse, and implicit functions.
In this section, we explore derivatives of exponential and logarithmic functions. As we discussed in Introduction to
Functions and Graphs, exponential functions play an important role in modeling population growth and the decay
of radioactive materials. Logarithmic functions can help rescale large quantities and are particularly helpful for rewriting
complicated expressions.
Derivative of the Exponential Function
Just as when we found the derivatives of other functions, we can find the derivatives of exponential and logarithmic
functions using formulas. As we develop these formulas, we need to make certain basic assumptions. The proofs that these
assumptions hold are beyond the scope of this course.
First of all, we begin with the assumption that the function B(x) = b x, b > 0, is defined for every real number and is
continuous. In previous courses, the values of exponential functions for all rational numbers were defined—beginning
with the definition of b n, where n is a positive integer—as the product of b multiplied by itself n times. Later,
we defined b 0 = 1, b −n = 1n ,
b
for a positive integer n,
t
and b s/t = ( b) s for positive integers s and t. These
definitions leave open the question of the value of b r where r is an arbitrary real number. By assuming the continuity of
B(x) = b x, b > 0, we may interpret b r as xlim
b x where the values of x as we take the limit are rational. For example,
→r
we may view 4 π as the number satisfying
4 3 < 4 π < 4 4, 4 3.1 < 4 π < 4 3.2, 4 3.14 < 4 π < 4 3.15,
4 3.141 < 4 π < 4 3.142, 4 3.1415 < 4 π < 4 3.1416 ,….
As we see in the following table, 4 π ≈ 77.88.
320
Chapter 3 | Derivatives
x
4x
x
4x
43
64
4 3.141593
77.8802710486
4 3.1
73.5166947198
4 3.1416
77.8810268071
4 3.14
77.7084726013
4 3.142
77.9242251944
4 3.141
77.8162741237
4 3.15
78.7932424541
4 3.1415
77.8702309526
4 3.2
84.4485062895
4 3.14159
77.8799471543
44
256
Table 3.7 Approximating a Value of 4 π
We also assume that for B(x) = b x, b > 0, the value B′ (0) of the derivative exists. In this section, we show that by
making this one additional assumption, it is possible to prove that the function B(x) is differentiable everywhere.
We make one final assumption: that there is a unique value of b > 0 for which B′ (0) = 1. We define e to be this
unique value, as we did in Introduction to Functions and Graphs. Figure 3.33 provides graphs of the functions
y = 2 x, y = 3 x, y = 2.7 x, and y = 2.8 x. A visual estimate of the slopes of the tangent lines to these functions at 0
provides evidence that the value of e lies somewhere between 2.7 and 2.8. The function E(x) = e x is called the natural
exponential function. Its inverse, L(x) = log e x = ln x is called the natural logarithmic function.
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Chapter 3 | Derivatives
321
Figure 3.33 The graph of E(x) = e x is between y = 2 x and y = 3 x.
For a better estimate of e, we may construct a table of estimates of B′ (0) for functions of the form B(x) = b x. Before
doing this, recall that
0
x
x
x
B′ (0) = lim b − b = lim b x− 1 ≈ b x− 1
x→0 x−0
x→0
for values of x very close to zero. For our estimates, we choose x = 0.00001 and x = −0.00001 to obtain the estimate
b −0.00001 − 1 < B′ (0) < b 0.00001 − 1 .
−0.00001
0.00001
See the following table.
322
Chapter 3 | Derivatives
b
b −0.00001 − 1 < B′ (0) < b 0.00001 − 1
−0.00001
0.00001
b
b −0.00001 − 1 < B′ (0) < b 0.00001 − 1
−0.00001
0.00001
2
0.693145 < B′ (0) < 0.69315
2.7183
1.000002 < B′ (0) < 1.000012
2.7
0.993247 < B′ (0) < 0.993257
2.719
1.000259 < B′ (0) < 1.000269
2.71
0.996944 < B′ (0) < 0.996954
2.72
1.000627 < B′ (0) < 1.000637
2.718
0.999891 < B′ (0) < 0.999901
2.8
1.029614 < B′ (0) < 1.029625
2.7182
0.999965 < B′ (0) < 0.999975
3
1.098606 < B′ (0) < 1.098618
Table 3.8 Estimating a Value of e
The evidence from the table suggests that 2.7182 < e < 2.7183.
The graph of E(x) = e x together with the line y = x + 1 are shown in Figure 3.34. This line is tangent to the graph of
E(x) = e x at x = 0.
Figure 3.34 The tangent line to E(x) = e x at x = 0 has
slope 1.
Now that we have laid out our basic assumptions, we begin our investigation by exploring the derivative of
B(x) = b x, b > 0. Recall that we have assumed that B′ (0) exists. By applying the limit definition to the derivative we
conclude that
0+h
− b 0 = lim b h − 1 .
B′ (0) = lim b
h
h→0
h→0 h
Turning to B′ (x), we obtain the following.
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Chapter 3 | Derivatives
323
x+h
− bx
B′ (x) = lim b
h
h→0
Apply the limit definition of he derivative.
x h
x
= lim b b − b
h
h→0
Note that b x + h = b x b h.
= lim
Factor out b x.
h
= b x lim b − 1
h→0 h
Apply a property of limits.
b x(b h − 1)
h
h→0
0+h
− b 0 = lim b h − 1 .
Use B′ (0) = lim b
h
h→0
h→0 h
= b x B′ (0)
We see that on the basis of the assumption that B(x) = b x is differentiable at 0, B(x) is not only differentiable everywhere,
but its derivative is
B′ (x) = b x B′ (0).
(3.29)
For E(x) = e x, E′ (0) = 1. Thus, we have E′ (x) = e x. (The value of B′ (0) for an arbitrary function of the form
B(x) = b x, b > 0, will be derived later.)
Theorem 3.14: Derivative of the Natural Exponential Function
Let E(x) = e x be the natural exponential function. Then
E′ (x) = e x.
In general,
d ⎛e g(x)⎞ = e g(x) g′ (x).
⎠
dx ⎝
Example 3.74
Derivative of an Exponential Function
Find the derivative of f (x) = e
tan(2x)
.
Solution
Using the derivative formula and the chain rule,
f ′ (x) = e
tan(2x)
=e
tan(2x)
Example 3.75
Combining Differentiation Rules
2
x
Find the derivative of y = ex .
d ⎛⎝tan(2x)⎞⎠
dx
sec 2 (2x) · 2.
324
Chapter 3 | Derivatives
Solution
Use the derivative of the natural exponential function, the quotient rule, and the chain rule.
⎛ x2 ⎞
x2
⎝e · 2⎠x · x − 1 · e
y′ =
x2
e x ⎛⎝2x 2 − 1⎞⎠
Apply the quotient rule.
2
=
3.50
Simplify.
x2
Find the derivative of h(x) = xe 2x.
Example 3.76
Applying the Natural Exponential Function
A colony of mosquitoes has an initial population of 1000. After t days, the population is given by
A(t) = 1000e 0.3t. Show that the ratio of the rate of change of the population, A′ (t), to the population, A(t) is
constant.
Solution
First find A′ (t). By using the chain rule, we have A′ (t) = 300e 0.3t. Thus, the ratio of the rate of change of the
population to the population is given by
0.3t
A′ (t) = 300e 0.3t = 0.3.
1000e
The ratio of the rate of change of the population to the population is the constant 0.3.
3.51
If A(t) = 1000e 0.3t describes the mosquito population after t days, as in the preceding example, what
is the rate of change of A(t) after 4 days?
Derivative of the Logarithmic Function
Now that we have the derivative of the natural exponential function, we can use implicit differentiation to find the derivative
of its inverse, the natural logarithmic function.
Theorem 3.15: The Derivative of the Natural Logarithmic Function
If x > 0 and y = ln x, then
dy 1
= .
dx x
(3.30)
More generally, let g(x) be a differentiable function. For all values of x for which g′ (x) > 0, the derivative of
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Chapter 3 | Derivatives
325
h(x) = ln ⎛⎝g(x)⎞⎠ is given by
h′ (x) = 1 g′ (x).
g(x)
(3.31)
Proof
If x > 0 and y = ln x, then e y = x. Differentiating both sides of this equation results in the equation
ey
Solving for
dy
= 1.
dx
dy
yields
dx
dy
= 1.
dx e y
Finally, we substitute x = e y to obtain
dy 1
= .
dx x
We may also derive this result by applying the inverse function theorem, as follows. Since y = g(x) = ln x is the inverse
of f (x) = e x, by applying the inverse function theorem we have
dy
1 = 1.
= ⎛ 1 ⎞ = ln
dx
f ′ ⎝g(x)⎠ e x x
Using this result and applying the chain rule to h(x) = ln ⎛⎝g(x)⎞⎠ yields
h′ (x) = 1 g′ (x).
g(x)
□
The graph of y = ln x and its derivative
dy 1
= are shown in Figure 3.35.
dx x
Figure 3.35
The function y = ln x is increasing on
(0, +∞). Its derivative y′ = 1x is greater than zero on
(0, +∞).
Example 3.77
Taking a Derivative of a Natural Logarithm
326
Chapter 3 | Derivatives
Find the derivative of f (x) = ln ⎛⎝x 3 + 3x − 4⎞⎠.
Solution
Use Equation 3.31 directly.
1
· ⎛⎝3x 2 + 3⎞⎠
x 3 + 3x − 4
2
= 33x + 3
x + 3x − 4
f ′ (x) =
Use g(x) = x 3 + 3x − 4 in h′ (x) = 1 g′ (x).
g(x)
Rewrite.
Example 3.78
Using Properties of Logarithms in a Derivative
⎞
⎛ 2
Find the derivative of f (x) = ln x sin x .
⎝ 2x + 1 ⎠
Solution
At first glance, taking this derivative appears rather complicated. However, by using the properties of logarithms
prior to finding the derivative, we can make the problem much simpler.
⎛ 2
⎞
f (x) = ln ⎝x sin x ⎠ = 2ln x + ln(sin x) − ln(2x + 1)
2x + 1
f ′ (x) = 2x + cot x − 2
2x + 1
3.52
Apply properties of logarithms.
Apply sum rule and h′ (x) = 1 g′ (x).
g(x)
Differentiate: f (x) = ln(3x + 2) 5.
Now that we can differentiate the natural logarithmic function, we can use this result to find the derivatives of y = log b x
and y = b x for b > 0, b ≠ 1.
Theorem 3.16: Derivatives of General Exponential and Logarithmic Functions
Let b > 0, b ≠ 1, and let g(x) be a differentiable function.
i. If, y = log b x, then
dy
= 1 .
dx x lnb
(3.32)
More generally, if h(x) = log b ⎛⎝g(x)⎞⎠, then for all values of x for which g(x) > 0,
h′ (x) =
g′ (x)
.
g(x)lnb
ii. If y = b x, then
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(3.33)
Chapter 3 | Derivatives
327
More generally, if h(x) = b
dy
= b x lnb.
dx
(3.34)
g(x)
(3.35)
g(x)
, then
h′ (x) = b
g″(x)lnb.
Proof
If y = log b x, then b y = x. It follows that ln(b y) = ln x. Thus y ln b = ln x. Solving for y, we have y = ln x .
lnb
Differentiating and keeping in mind that lnb is a constant, we see that
dy
= 1 .
dx x lnb
The derivative in Equation 3.33 now follows from the chain rule.
If y = b x, then ln y = x lnb. Using implicit differentiation, again keeping in mind that lnb is constant, it follows that
1 dy = lnb. Solving for dy and substituting y = b x, we see that
y dx
dx
dy
= ylnb = b x lnb.
dx
The more general derivative (Equation 3.35) follows from the chain rule.
□
Example 3.79
Applying Derivative Formulas
x
Find the derivative of h(x) = x3 .
3 +2
Solution
Use the quotient rule and Derivatives of General Exponential and Logarithmic Functions.
3 x ln3(3 x + 2) − 3 x ln3(3 x)
(3 x + 2) 2
x
= 2 ·x3 ln32
(3 + 2)
h′ (x) =
Apply the quotient rule.
Simplify.
Example 3.80
Finding the Slope of a Tangent Line
Find the slope of the line tangent to the graph of y = log 2 (3x + 1) at x = 1.
Solution
328
Chapter 3 | Derivatives
To find the slope, we must evaluate
dy
at x = 1. Using Equation 3.33, we see that
dx
dy
3
=
.
dx ln2(3x + 1)
By evaluating the derivative at x = 1, we see that the tangent line has slope
dy
dx
3.53
|
x=1
=
3 = 3 .
4ln2 ln16
Find the slope for the line tangent to y = 3 x at x = 2.
Logarithmic Differentiation
At this point, we can take derivatives of functions of the form y = ⎛⎝g(x)⎞⎠ n for certain values of n, as well as functions
of the form y = b
x
g(x)
π
, where b > 0 and b ≠ 1. Unfortunately, we still do not know the derivatives of functions such as
y = x or y = x . These functions require a technique called logarithmic differentiation, which allows us to differentiate
any function of the form h(x) = g(x)
f (x)
. It can also be used to convert a very complex differentiation problem into a
simpler one, such as finding the derivative of y = x x2x +3 1 . We outline this technique in the following problem-solving
e sin x
strategy.
Problem-Solving Strategy: Using Logarithmic Differentiation
1. To differentiate y = h(x) using logarithmic differentiation, take the natural logarithm of both sides of the
equation to obtain ln y = ln ⎛⎝h(x)⎞⎠.
2. Use properties of logarithms to expand ln ⎛⎝h(x)⎞⎠ as much as possible.
dy
3. Differentiate both sides of the equation. On the left we will have 1y .
dx
4. Multiply both sides of the equation by y to solve for
5. Replace y by h(x).
Example 3.81
Using Logarithmic Differentiation
Find the derivative of y = ⎛⎝2x 4 + 1⎞⎠
tan x
.
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dy
.
dx
Chapter 3 | Derivatives
329
Solution
Use logarithmic differentiation to find this derivative.
ln y = ln⎛⎝2x 4 + 1⎞⎠
ln y =
tan x
tan x ln ⎛⎝2x 4
+
Step 1. Take the natural logarithm of both sides.
1⎞⎠
Step 2. Expand using properties of logarithms.
Step 3. Diffe entiate both sides. Use the
product rule on the right.
1 dy = sec 2 x ln ⎛2x 4 + 1⎞ + 8x 3 · tan x
⎝
⎠
y dx
2x 4 + 1
3
⎛
⎞
dy
= y · sec 2 x ln ⎛⎝2x 4 + 1⎞⎠ + 8x
· tan x
4
⎝
⎠
dx
2x + 1
dy
=
dx
⎛ 4
⎝2x
+ 1⎞⎠
tan x ⎛
sec 2 x ln ⎛⎝2x 4 + 1⎞⎠ +
⎝
8x 3 · tan x⎞
⎠
2x 4 + 1
Step 4. Multiply by y on both sides.
Step 5. Substitute y = ⎛⎝2x 4 + 1⎞⎠
tan x
.
Example 3.82
Using Logarithmic Differentiation
Find the derivative of y = x x2x +3 1 .
e sin x
Solution
This problem really makes use of the properties of logarithms and the differentiation rules given in this chapter.
ln y = ln x x2x +3 1
e sin x
ln y = ln x + 1 ln(2x + 1) − x lne − 3lnsin x
2
1 dy = 1 + 1 − 1 − 3 cos x
x 2x + 1
y dx
sin x
dy
⎛1
⎞
= y⎝ x + 1 − 1 − 3cot x⎠
2x + 1
dx
dy
⎛
⎞
= x x2x +3 1 ⎝1x + 1 − 1 − 3cot x⎠
2x
+
1
dx
e sin x
Step 1. Take the natural logarithm of both sides.
Step 2. Expand using properties of logarithms.
Step 3. Diffe entiate both sides.
Step 4. Multiply by y on both sides.
Step 5. Substitute y = x x2x +3 1 .
e sin x
Example 3.83
Extending the Power Rule
Find the derivative of y = x r where r is an arbitrary real number.
Solution
The process is the same as in Example 3.82, though with fewer complications.
330
Chapter 3 | Derivatives
ln y
ln y
1 dy
y dx
dy
dx
dy
dx
dy
dx
= lnx r
= r ln x
= r 1x
Step 1. Take the natural logarithm of both sides.
Step 2. Expand using properties of logarithms.
= y xr
Step 4. Multiply by y on both sides.
= x r xr
Step 5. Substitute y = x r.
= rx r − 1
Simplify.
Step 3. Diffe entiate both sides.
3.54
Use logarithmic differentiation to find the derivative of y = x x.
3.55
Find the derivative of y = (tan x) π.
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Chapter 3 | Derivatives
331
3.9 EXERCISES
For the following exercises, find f ′ (x) for each function.
331.
f (x) = x 2 e x
332.
−x
f (x) = e x
333.
f (x) = e x
334.
f (x) = e 2x + 2x
335.
−x
x
f (x) = e x − e −x
e +e
336.
x
f (x) = 10
ln10
350. y = ⎛⎝x 2 − 1⎞⎠
ln x
351. y = x cot x
352. y = 3x + 11
x2 − 4
3 ln x
2/3
353. y = x −1/2 ⎛⎝x 2 + 3⎞⎠ (3x − 4) 4
354. [T] Find an equation of the tangent line to the graph
of f (x) = 4xe
⎛ 2
⎞
⎝x − 1⎠
at the point where x = −1. Graph
both the function and the tangent line.
355. [T] Find the equation of the line that is normal to the
graph of f (x) = x · 5 x at the point where x = 1. Graph
337.
f (x) = 2 4x + 4x 2
338.
f (x) = 3 sin3x
339.
f (x) = x π · π x
340.
f (x) = ln ⎛⎝4x 3 + x⎞⎠
the curve and the tangent line.
341.
f (x) = ln 5x − 7
357. Consider the function y = x 1/x for x > 0.
342.
2
f (x) = x ln9x
343.
f (x) = log(sec x)
344.
f (x) = log 7 ⎛⎝6x 4 + 3⎞⎠
345.
f (x) = 2 x · log 3 7 x
both the function and the normal line.
356. [T] Find the equation of the tangent line to the graph
of x 3 − x ln y + y 3 = 2x + 5 at the point where x = 2.
(Hint: Use implicit differentiation to find
a. Determine the points on the graph where the
tangent line is horizontal.
b. Determine the points on the graph where y′ > 0
and those where y′ < 0.
5
2−4
For the following exercises, use logarithmic differentiation
to find
dy
.
dx
346. y = x x
347. y = (sin2x) 4x
348. y = (ln x) ln x
349. y = x
log 2 x
dy
.) Graph both
dx
332
Chapter 3 | Derivatives
358.
The formula I(t) = sint
is the formula for a
t
e
decaying alternating current.
a. Complete the following table with the appropriate
values.
t
sint
et
360. [T] An isotope of the element erbium has a half-life
of approximately 12 hours. Initially there are 9 grams of the
isotope present.
a. Write the exponential function that relates the
amount of substance remaining as a function of t,
measured in hours.
b. Use a. to determine the rate at which the substance
is decaying in t hours.
c. Use b. to determine the rate of decay at t = 4
hours.
0
(i)
π
2
(ii)
π
(iii)
N(t) = 5.3e 0.093t
3π
2
(iv)
gives the number of cases (in thousands) and t is measured
in years, with t = 0 corresponding to the beginning of
1960.
a. Show work that evaluates N(0) and N(4). Briefly
2π
(v)
describe what these values indicate about the
disease in New York City.
b. Show work that evaluates N′ (0) and N′ (3).
2π
(vi)
Briefly describe what these values indicate about
the disease in the United States.
3π
(vii)
7π
2
(viii)
4π
(ix)
b. Using only the values in the table, determine where
the tangent line to the graph of I(t) is horizontal.
359. [T] The population of Toledo, Ohio, in 2000 was
approximately 500,000. Assume the population is
increasing at a rate of 5% per year.
a. Write the exponential function that relates the total
population as a function of t.
b. Use a. to determine the rate at which the population
is increasing in t years.
c. Use b. to determine the rate at which the population
is increasing in 10 years.
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361. [T] The number of cases of influenza in New York
City from the beginning of 1960 to the beginning of 1961
is
modeled
by
the
function
2 − 0.87t
, (0 ≤ t ≤ 4),
where
N(t)
362. [T] The relative rate of change of a differentiable
function y = f (x) is given by
100 · f ′ (x)
%. One model
f (x)
for population growth is a Gompertz growth function,
given by P(x) = ae −b · e
−cx
where a, b,
and c are
constants.
a. Find the relative rate of change formula for the
generic Gompertz function.
b. Use a. to find the relative rate of change of a
population
in
x = 20
months
when
a = 204, b = 0.0198, and c = 0.15.
c. Briefly interpret what the result of b. means.
For the following exercises, use the population of New
York City from 1790 to 1860, given in the following table.
Chapter 3 | Derivatives
333
Years since 1790
Population
0
33,131
10
60,515
20
96,373
30
123,706
40
202,300
50
312,710
60
515,547
70
813,669
Table 3.9 New York City Population Over
Time Source: http://en.wikipedia.org/
wiki/
Largest_cities_in_the_United_States
_by_population_by_decade.
363. [T] Using a computer program or a calculator, fit a
growth curve to the data of the form p = ab t.
364. [T] Using the exponential best fit for the data, write
a table containing the derivatives evaluated at each year.
365. [T] Using the exponential best fit for the data, write
a table containing the second derivatives evaluated at each
year.
366. [T] Using the tables of first and second derivatives
and the best fit, answer the following questions:
a. Will the model be accurate in predicting the future
population of New York City? Why or why not?
b. Estimate the population in 2010. Was the prediction
correct from a.?
334
Chapter 3 | Derivatives
CHAPTER 3 REVIEW
KEY TERMS
acceleration is the rate of change of the velocity, that is, the derivative of velocity
amount of change the amount of a function f (x) over an interval ⎡⎣x, x + h⎤⎦ is f (x + h) − f (x)
average rate of change
is a function f (x) over an interval ⎡⎣x, x + h⎤⎦ is
f (x + h) − f (a)
b−a
chain rule the chain rule defines the derivative of a composite function as the derivative of the outer function evaluated
at the inner function times the derivative of the inner function
constant multiple rule the derivative of a constant c multiplied by a function f is the same as the constant multiplied by
the derivative: d ⎛⎝c f (x)⎞⎠ = c f ′ (x)
dx
constant rule the derivative of a constant function is zero: d (c) = 0, where c is a constant
dx
derivative the slope of the tangent line to a function at a point, calculated by taking the limit of the difference quotient, is
the derivative
derivative function gives the derivative of a function at each point in the domain of the original function for which the
derivative is defined
difference quotient of a function f (x) at a is given by
f (x) − f (a)
f (a + h) − f (a)
or
x−a
h
difference rule the derivative of the difference of a function f and a function g is the same as the difference of the
derivative of f and the derivative of g: d ⎛⎝ f (x) − g(x)⎞⎠ = f ′ (x) − g′ (x)
dx
differentiable at a a function for which f ′(a) exists is differentiable at a
differentiable function a function for which f ′(x) exists is a differentiable function
differentiable on S a function for which f ′(x) exists for each x in the open set S is differentiable on S
differentiation the process of taking a derivative
higher-order derivative a derivative of a derivative, from the second derivative to the nth derivative, is called a higherorder derivative
dy
for a function defined by an equation, accomplished by
dx
dy
differentiating both sides of the equation (remembering to treat the variable y as a function) and solving for
dx
implicit differentiation
is a technique for computing
instantaneous rate of change the rate of change of a function at any point along the function a, also called f ′(a),
or the derivative of the function at a
logarithmic differentiation is a technique that allows us to differentiate a function by first taking the natural logarithm
of both sides of an equation, applying properties of logarithms to simplify the equation, and differentiating implicitly
marginal cost is the derivative of the cost function, or the approximate cost of producing one more item
marginal profit is the derivative of the profit function, or the approximate profit obtained by producing and selling one
more item
marginal revenue is the derivative of the revenue function, or the approximate revenue obtained by selling one more
item
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Chapter 3 | Derivatives
335
population growth rate is the derivative of the population with respect to time
power rule the derivative of a power function is a function in which the power on x becomes the coefficient of the term
and the power on x in the derivative decreases by 1: If n is an integer, then d x n = nx n − 1
dx
product rule the derivative of a product of two functions is the derivative of the first function times the second function
plus the derivative of the second function times the first function: d ⎛⎝ f (x)g(x)⎞⎠ = f ′ (x)g(x) + g′ (x) f (x)
dx
quotient rule the derivative of the quotient of two functions is the derivative of the first function times the second
function minus the derivative of the second function times the first function, all divided by the square of the second
⎛ f (x) ⎞ f ′ (x)g(x) − g′ (x) f (x)
=
function: d ⎝
⎞2
⎛
dx g(x) ⎠
⎝g(x)⎠
speed is the absolute value of velocity, that is, |v(t)| is the speed of an object at time t whose velocity is given by v(t)
sum rule the derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the
derivative of g: d ⎛⎝ f (x) + g(x)⎞⎠ = f ′ (x) + g′ (x)
dx
KEY EQUATIONS
• Difference quotient
Q=
f (x) − f (a)
x−a
• Difference quotient with increment h
Q=
f (a + h) − f (a)
f (a + h) − f (a)
=
a+h−a
h
• Slope of tangent line
f (x) − f (a)
x−a
f (a + h) − f (a)
m tan = lim
h
h→0
m tan = xlim
→a
• Derivative of f (x) at a
f (x) − f (a)
x−a
f (a + h) − f (a)
f ′(a) = lim
h
h→0
f ′ (a) = xlim
→a
• Average velocity
v ave =
s(t) − s(a)
t−a
• Instantaneous velocity
v(a) = s′ (a) = lim
t→a
s(t) − s(a)
t−a
• The derivative function
f ′ (x) = lim
h→0
f (x + h) − f (x)
h
• Derivative of sine function
d (sin x) = cos x
dx
• Derivative of cosine function
d (cos x) = −sin x
dx
• Derivative of tangent function
336
Chapter 3 | Derivatives
d (tan x) = sec 2 x
dx
• Derivative of cotangent function
d (cot x) = −csc 2 x
dx
• Derivative of secant function
d (sec x) = sec x tan x
dx
• Derivative of cosecant function
d (csc x) = −csc x cot x
dx
• The chain rule
h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x)
• The power rule for functions
h′ (x) = n⎛⎝g(x)⎞⎠ n − 1 g′ (x)
• Inverse function theorem
⎛ −1⎞
⎝f
⎠′ (x)
=
1
whenever f ′ ⎛⎝ f −1 (x)⎞⎠ ≠ 0 and f (x) is differentiable.
f ′ ⎛⎝ f −1 (x)⎞⎠
• Power rule with rational exponents
d ⎛x m/n⎞ = m x (m/n) − 1.
⎠
n
dx ⎝
• Derivative of inverse sine function
d sin −1 x =
1
dx
1 − (x) 2
• Derivative of inverse cosine function
d cos −1 x =
−1
dx
1 − (x) 2
• Derivative of inverse tangent function
d tan −1 x =
1
dx
1 + (x) 2
• Derivative of inverse cotangent function
d cot −1 x = −1
dx
1 + (x) 2
• Derivative of inverse secant function
d sec −1 x =
1
dx
|x| (x) 2 − 1
• Derivative of inverse cosecant function
d csc −1 x =
−1
dx
|x| (x) 2 − 1
• Derivative of the natural exponential function
d ⎛e g(x)⎞ = e g(x) g′ (x)
⎠
dx ⎝
• Derivative of the natural logarithmic function
d ⎛⎝lng(x)⎞⎠ = 1 g′ (x)
g(x)
dx
• Derivative of the general exponential function
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Chapter 3 | Derivatives
337
d ⎛b g(x)⎞ = b g(x) g′ (x)lnb
⎠
dx ⎝
• Derivative of the general logarithmic function
d ⎛log g(x)⎞⎠ = g′ (x)
g(x)lnb
dx ⎝ b
KEY CONCEPTS
3.1 Defining the Derivative
• The slope of the tangent line to a curve measures the instantaneous rate of change of a curve. We can calculate it by
finding the limit of the difference quotient or the difference quotient with increment h.
• The derivative of a function f (x) at a value a is found using either of the definitions for the slope of the tangent
line.
• Velocity is the rate of change of position. As such, the velocity v(t) at time t is the derivative of the position s(t)
at time t. Average velocity is given by
v ave =
s(t) − s(a)
t−a .
Instantaneous velocity is given by
v(a) = s′ (a) = lim
t→a
s(t) − s(a)
t−a .
• We may estimate a derivative by using a table of values.
3.2 The Derivative as a Function
• The derivative of a function f (x) is the function whose value at x is f ′(x).
• The graph of a derivative of a function f (x) is related to the graph of f (x). Where f (x) has a tangent line with
positive slope, f ′ (x) > 0. Where f (x) has a tangent line with negative slope, f ′ (x) < 0. Where f (x) has a
horizontal tangent line, f ′ (x) = 0.
• If a function is differentiable at a point, then it is continuous at that point. A function is not differentiable at a point
if it is not continuous at the point, if it has a vertical tangent line at the point, or if the graph has a sharp corner or
cusp.
• Higher-order derivatives are derivatives of derivatives, from the second derivative to the nth derivative.
3.3 Differentiation Rules
• The derivative of a constant function is zero.
• The derivative of a power function is a function in which the power on x becomes the coefficient of the term and
the power on x in the derivative decreases by 1.
• The derivative of a constant c multiplied by a function f is the same as the constant multiplied by the derivative.
• The derivative of the sum of a function f and a function g is the same as the sum of the derivative of f and the
derivative of g.
• The derivative of the difference of a function f and a function g is the same as the difference of the derivative of f
and the derivative of g.
• The derivative of a product of two functions is the derivative of the first function times the second function plus the
derivative of the second function times the first function.
• The derivative of the quotient of two functions is the derivative of the first function times the second function minus
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Chapter 3 | Derivatives
the derivative of the second function times the first function, all divided by the square of the second function.
• We used the limit definition of the derivative to develop formulas that allow us to find derivatives without resorting
to the definition of the derivative. These formulas can be used singly or in combination with each other.
3.4 Derivatives as Rates of Change
• Using f (a + h) ≈ f (a) + f ′ (a)h, it is possible to estimate f (a + h) given f ′ (a) and f (a).
• The rate of change of position is velocity, and the rate of change of velocity is acceleration. Speed is the absolute
value, or magnitude, of velocity.
• The population growth rate and the present population can be used to predict the size of a future population.
• Marginal cost, marginal revenue, and marginal profit functions can be used to predict, respectively, the cost of
producing one more item, the revenue obtained by selling one more item, and the profit obtained by producing and
selling one more item.
3.5 Derivatives of Trigonometric Functions
• We can find the derivatives of sin x and cos x by using the definition of derivative and the limit formulas found
earlier. The results are
d sin x = cos x d cos x = −sin x.
dx
dx
• With these two formulas, we can determine the derivatives of all six basic trigonometric functions.
3.6 The Chain Rule
• The chain rule allows us to differentiate compositions of two or more functions. It states that for h(x) = f ⎛⎝g(x)⎞⎠,
h′ (x) = f ′ ⎛⎝g(x)⎞⎠g′ (x).
In Leibniz’s notation this rule takes the form
dy dy du
=
· .
dx du dx
• We can use the chain rule with other rules that we have learned, and we can derive formulas for some of them.
• The chain rule combines with the power rule to form a new rule:
If h(x) = ⎛⎝g(x)⎞⎠ n, then h′ (x) = n⎛⎝g(x)⎞⎠ n − 1 g′ (x).
• When applied to the composition of three functions, the chain rule can be expressed as follows: If
h(x) = f ⎛⎝g⎛⎝k(x)⎞⎠⎞⎠, then h′ (x) = f ′(g⎛⎝k(x)⎞⎠g′ ⎛⎝k(x)⎞⎠k′ (x).
3.7 Derivatives of Inverse Functions
• The inverse function theorem allows us to compute derivatives of inverse functions without using the limit
definition of the derivative.
• We can use the inverse function theorem to develop differentiation formulas for the inverse trigonometric functions.
3.8 Implicit Differentiation
• We use implicit differentiation to find derivatives of implicitly defined functions (functions defined by equations).
• By using implicit differentiation, we can find the equation of a tangent line to the graph of a curve.
3.9 Derivatives of Exponential and Logarithmic Functions
• On the basis of the assumption that the exponential function y = b x, b > 0 is continuous everywhere and
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Chapter 3 | Derivatives
339
differentiable at 0, this function is differentiable everywhere and there is a formula for its derivative.
• We can use a formula to find the derivative of y = ln x, and the relationship log b x = ln x allows us to extend
lnb
our differentiation formulas to include logarithms with arbitrary bases.
• Logarithmic differentiation allows us to differentiate functions of the form y = g(x)
f (x)
or very complex functions
by taking the natural logarithm of both sides and exploiting the properties of logarithms before differentiating.
CHAPTER 3 REVIEW EXERCISES
True or False? Justify the answer with a proof or a
counterexample.
382. Third derivative of y = (3x + 2) 2
367. Every function has a derivative.
383. Second derivative of y = 4 x + x 2 sin(x)
368. A continuous function has a continuous derivative.
369. A continuous function has a derivative.
370. If a function is differentiable, it is continuous.
Use the limit definition of the derivative to exactly evaluate
the derivative.
371.
f (x) = x + 4
372.
f (x) = 3x
Find the equation of the tangent line to the following
equations at the specified point.
384. y = cos −1 (x) + x at x = 0
385. y = x + e x − 1x at x = 1
Draw the derivative for the following graphs.
386.
Find the derivatives of the following functions.
373.
f (x) = 3x 3 − 42
x
374.
f (x) = ⎛⎝4 − x 2⎞⎠
375.
f (x) = e sin x
376.
f (x) = ln(x + 2)
377.
f (x) = x 2 cos x + x tan(x)
378.
f (x) = 3x 2 + 2
379.
f (x) = x sin −1 (x)
4
2
⎛
⎝
3
⎞
⎠
380. x y = y + 2 + xysin(x)
Find the following derivatives of various orders.
381. First derivative of y = x ln(x)cos x
387.
The following questions concern the water level in Ocean
City, New Jersey, in January, which can be approximated
⎛
⎞
by w(t) = 1.9 + 2.9cos ⎝π t⎠,
6
where t is measured in
hours after midnight, and the height is measured in feet.
388. Find and graph the derivative. What is the physical
meaning?
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Chapter 3 | Derivatives
389. Find w′ (3). What is the physical meaning of this
value?
The following questions consider the wind speeds of
Hurricane Katrina, which affected New Orleans, Louisiana,
in August 2005. The data are displayed in a table.
Hours after Midnight,
August 26
Wind Speed
(mph)
1
45
5
75
11
100
29
115
49
145
58
175
73
155
81
125
85
95
107
35
Table 3.10 Wind Speeds of Hurricane
Katrina Source:
http://news.nationalgeographic.com/news/2005/
09/0914_050914_katrina_timeline.html.
390. Using the table, estimate the derivative of the wind
speed at hour 39. What is the physical meaning?
391. Estimate the derivative of the wind speed at hour 83.
What is the physical meaning?
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4 | APPLICATIONS OF
DERIVATIVES
Figure 4.1 As a rocket is being launched, at what rate should the angle of a video camera change to continue viewing the
rocket? (credit: modification of work by Steve Jurvetson, Wikimedia Commons)
Chapter Outline
4.1 Related Rates
4.2 Linear Approximations and Differentials
4.3 Maxima and Minima
4.4 The Mean Value Theorem
4.5 Derivatives and the Shape of a Graph
4.6 Limits at Infinity and Asymptotes
4.7 Applied Optimization Problems
4.8 L’Hôpital’s Rule
4.9 Newton’s Method
4.10 Antiderivatives
Introduction
A rocket is being launched from the ground and cameras are recording the event. A video camera is located on the ground
a certain distance from the launch pad. At what rate should the angle of inclination (the angle the camera makes with the
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Chapter 4 | Applications of Derivatives
ground) change to allow the camera to record the flight of the rocket as it heads upward? (See Example 4.3.)
A rocket launch involves two related quantities that change over time. Being able to solve this type of problem is just
one application of derivatives introduced in this chapter. We also look at how derivatives are used to find maximum and
minimum values of functions. As a result, we will be able to solve applied optimization problems, such as maximizing
revenue and minimizing surface area. In addition, we examine how derivatives are used to evaluate complicated limits, to
approximate roots of functions, and to provide accurate graphs of functions.
4.1 | Related Rates
Learning Objectives
4.1.1 Express changing quantities in terms of derivatives.
4.1.2 Find relationships among the derivatives in a given problem.
4.1.3 Use the chain rule to find the rate of change of one quantity that depends on the rate of
change of other quantities.
We have seen that for quantities that are changing over time, the rates at which these quantities change are given by
derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For
example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing.
In this section, we consider several problems in which two or more related quantities are changing and we study how to
determine the relationship between the rates of change of these quantities.
Setting up Related-Rates Problems
In many real-world applications, related quantities are changing with respect to time. For example, if we consider the
balloon example again, we can say that the rate of change in the volume, V, is related to the rate of change in the radius,
r. In this case, we say that dV and dr are related rates because V is related to r. Here we study several examples of
dt
dt
related quantities that are changing with respect to time and we look at how to calculate one rate of change given another
rate of change.
Example 4.1
Inflating a Balloon
A spherical balloon is being filled with air at the constant rate of 2 cm 3 /sec (Figure 4.2). How fast is the radius
increasing when the radius is 3 cm ?
Figure 4.2 As the balloon is being filled with air, both the radius and the volume are increasing with respect to time.
Solution
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Chapter 4 | Applications of Derivatives
343
The volume of a sphere of radius r centimeters is
V = 4 πr 3 cm 3.
3
Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, t
seconds after beginning to fill the balloon with air, the volume of air in the balloon is
V(t) = 4 π[r(t)] 3 cm 3.
3
Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of
change in the volume is related to the rate of change in the radius by the equation
V′(t) = 4π ⎡⎣r(t)⎤⎦ 2 r′ (t).
The balloon is being filled with air at the constant rate of 2 cm3/sec, so V′(t) = 2 cm 3 /sec. Therefore,
2cm 3 /sec = ⎛⎝4π ⎡⎣r(t)⎤⎦ 2 cm 2⎞⎠ · ⎛⎝r′(t)cm/s⎞⎠,
which implies
r′(t) =
1
cm/sec.
2π ⎡⎣r(t)⎤⎦ 2
When the radius r = 3 cm,
r′(t) = 1 cm/sec.
18π
4.1
What is the instantaneous rate of change of the radius when r = 6 cm ?
Before looking at other examples, let’s outline the problem-solving strategy we will be using to solve related-rates problems.
Problem-Solving Strategy: Solving a Related-Rates Problem
1. Assign symbols to all variables involved in the problem. Draw a figure if applicable.
2. State, in terms of the variables, the information that is given and the rate to be determined.
3. Find an equation relating the variables introduced in step 1.
4. Using the chain rule, differentiate both sides of the equation found in step 3 with respect to the independent
variable. This new equation will relate the derivatives.
5. Substitute all known values into the equation from step 4, then solve for the unknown rate of change.
Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the
value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that
quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this
potential error in the following example.
Examples of the Process
Let’s now implement the strategy just described to solve several related-rates problems. The first example involves a plane
flying overhead. The relationship we are studying is between the speed of the plane and the rate at which the distance
between the plane and a person on the ground is changing.
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Chapter 4 | Applications of Derivatives
Example 4.2
An Airplane Flying at a Constant Elevation
An airplane is flying overhead at a constant elevation of 4000 ft. A man is viewing the plane from a position
3000 ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is
flying at the rate of 600 ft/sec, at what rate is the distance between the man and the plane increasing when the
plane passes over the radio tower?
Solution
Step 1. Draw a picture, introducing variables to represent the different quantities involved.
Figure 4.3 An airplane is flying at a constant height of 4000 ft. The distance between the
person and the airplane and the person and the place on the ground directly below the airplane
are changing. We denote those quantities with the variables s and x, respectively.
As shown, x denotes the distance between the man and the position on the ground directly below the airplane.
The variable s denotes the distance between the man and the plane. Note that both x and s are functions of
time. We do not introduce a variable for the height of the plane because it remains at a constant elevation of
4000 ft. Since an object’s height above the ground is measured as the shortest distance between the object and
the ground, the line segment of length 4000 ft is perpendicular to the line segment of length x feet, creating a
right triangle.
Step 2. Since x denotes the horizontal distance between the man and the point on the ground below the plane,
dx/dt represents the speed of the plane. We are told the speed of the plane is 600 ft/sec. Therefore, dx = 600
dt
ft/sec. Since we are asked to find the rate of change in the distance between the man and the plane when the plane
is directly above the radio tower, we need to find ds/dt when x = 3000 ft.
Step 3. From the figure, we can use the Pythagorean theorem to write an equation relating x and s:
⎡
⎣
x(t)⎤⎦ 2 + 4000 2 = ⎡⎣s(t)⎤⎦ 2.
Step 4. Differentiating this equation with respect to time and using the fact that the derivative of a constant is
zero, we arrive at the equation
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345
x dx = s ds .
dt
dt
Step 5. Find the rate at which the distance between the man and the plane is increasing when the plane is directly
over the radio tower. That is, find ds when x = 3000 ft. Since the speed of the plane is 600 ft/sec, we know
dt
that dx = 600 ft/sec. We are not given an explicit value for s; however, since we are trying to find ds when
dt
dt
x = 3000 ft, we can use the Pythagorean theorem to determine the distance s when x = 3000 and the height
is 4000 ft. Solving the equation
3000 2 + 4000 2 = s 2
for s, we have s = 5000 ft at the time of interest. Using these values, we conclude that ds/dt is a solution of
the equation
(3000)(600) = (5000) · ds .
dt
Therefore,
ds = 3000 · 600 = 360 ft/sec.
dt
5000
Note: When solving related-rates problems, it is important not to substitute values for the variables too soon. For
example, in step 3, we related the variable quantities x(t) and s(t) by the equation
⎡
⎣
x(t)⎤⎦ 2 + 4000 2 = ⎡⎣s(t)⎤⎦ 2.
Since the plane remains at a constant height, it is not necessary to introduce a variable for the height, and we are
allowed to use the constant 4000 to denote that quantity. However, the other two quantities are changing. If we
mistakenly substituted x(t) = 3000 into the equation before differentiating, our equation would have been
3000 2 + 4000 2 = ⎡⎣s(t)⎤⎦ 2.
After differentiating, our equation would become
0 = s(t) ds .
dt
As a result, we would incorrectly conclude that ds = 0.
dt
4.2 What is the speed of the plane if the distance between the person and the plane is increasing at the rate of
300 ft/sec?
We now return to the problem involving the rocket launch from the beginning of the chapter.
Example 4.3
Chapter Opener: A Rocket Launch
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Chapter 4 | Applications of Derivatives
Figure 4.4 (credit: modification of work by Steve Jurvetson,
Wikimedia Commons)
A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the
rocket is 1000 ft above the launch pad, its velocity is 600 ft/sec. Find the necessary rate of change of the
camera’s angle as a function of time so that it stays focused on the rocket.
Solution
Step 1. Draw a picture introducing the variables.
Figure 4.5 A camera is positioned 5000 ft from the launch pad of the rocket. The height of the
rocket and the angle of the camera are changing with respect to time. We denote those quantities
with the variables h and θ, respectively.
Let h denote the height of the rocket above the launch pad and θ be the angle between the camera lens and the
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Chapter 4 | Applications of Derivatives
347
ground.
Step 2. We are trying to find the rate of change in the angle of the camera with respect to time when the rocket is
1000 ft off the ground. That is, we need to find dθ when h = 1000 ft. At that time, we know the velocity of the
dt
rocket is dh = 600 ft/sec.
dt
Step 3. Now we need to find an equation relating the two quantities that are changing with respect to time: h and
θ. How can we create such an equation? Using the fact that we have drawn a right triangle, it is natural to think
about trigonometric functions. Recall that tanθ is the ratio of the length of the opposite side of the triangle to the
length of the adjacent side. Thus, we have
tanθ =
h .
5000
This gives us the equation
h = 5000tanθ.
Step 4. Differentiating this equation with respect to time t, we obtain
dh = 5000sec 2 θ dθ .
dt
dt
Step 5. We want to find dθ when h = 1000 ft. At this time, we know that dh = 600 ft/sec. We need to
dt
dt
2
determine sec θ. Recall that secθ is the ratio of the length of the hypotenuse to the length of the adjacent
side. We know the length of the adjacent side is 5000 ft. To determine the length of the hypotenuse, we use the
Pythagorean theorem, where the length of one leg is 5000 ft, the length of the other leg is h = 1000 ft, and
the length of the hypotenuse is c feet as shown in the following figure.
We see that
1000 2 + 5000 2 = c 2
and we conclude that the hypotenuse is
c = 1000 26 ft.
Therefore, when h = 1000, we have
⎛
⎞
sec 2 θ = ⎝1000 26 ⎠ = 26 .
5000
25
2
Recall from step 4 that the equation relating dθ to our known values is
dt
dh = 5000sec 2 θ dθ .
dt
dt
When h = 1000 ft,
we know that dh = 600 ft/sec and sec 2 θ = 26 . Substituting these values into the
dt
25
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Chapter 4 | Applications of Derivatives
previous equation, we arrive at the equation
⎛
⎞
600 = 5000⎝26 ⎠dθ .
25 dt
Therefore, dθ = 3 rad/sec.
dt
26
4.3 What rate of change is necessary for the elevation angle of the camera if the camera is placed on the
ground at a distance of 4000 ft from the launch pad and the velocity of the rocket is 500 ft/sec when the rocket
is 2000 ft off the ground?
In the next example, we consider water draining from a cone-shaped funnel. We compare the rate at which the level of water
in the cone is decreasing with the rate at which the volume of water is decreasing.
Example 4.4
Water Draining from a Funnel
Water is draining from the bottom of a cone-shaped funnel at the rate of 0.03 ft 3 /sec. The height of the funnel
is 2 ft and the radius at the top of the funnel is 1 ft. At what rate is the height of the water in the funnel changing
when the height of the water is 1 ft ?
2
Solution
Step 1: Draw a picture introducing the variables.
Figure 4.6 Water is draining from a funnel of height 2 ft and
radius 1 ft. The height of the water and the radius of water are
changing over time. We denote these quantities with the
variables h and r, respectively.
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349
Let h denote the height of the water in the funnel, r denote the radius of the water at its surface, and V denote
the volume of the water.
Step 2: We need to determine dh when h = 1 ft. We know that dV = −0.03 ft/sec.
2
dt
dt
Step 3: The volume of water in the cone is
V = 1 πr 2 h.
3
From the figure, we see that we have similar triangles. Therefore, the ratio of the sides in the two triangles is the
same. Therefore, r = 1 or r = h . Using this fact, the equation for volume can be simplified to
h
2
2
⎛ ⎞
V = 1 π ⎝h ⎠ h = π h 3.
3 2
12
2
Step 4: Applying the chain rule while differentiating both sides of this equation with respect to time t, we obtain
dV = π h 2 dh .
4 dt
dt
Step 5: We want to find dh when h = 1 ft. Since water is leaving at the rate of 0.03 ft 3 /sec, we know that
dt
dV = −0.03 ft 3 /sec. Therefore,
dt
2
⎛ ⎞
−0.03 = π ⎝1 ⎠ dh ,
4 2 dt
2
which implies
−0.03 = π dh .
16 dt
It follows that
dh = − 0.48 = −0.153 ft/sec.
π
dt
4.4
At what rate is the height of the water changing when the height of the water is 1 ft ?
4
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Chapter 4 | Applications of Derivatives
4.1 EXERCISES
For the following exercises, find the quantities for the given
equation.
1. Find
dy
at x = 1 and y = x 2 + 3 if dx = 4.
dt
dt
dy
2. Find dx at x = −2 and y = 2x 2 + 1 if
= −1.
dt
3.
7. Two airplanes are flying in the air at the same height:
airplane A is flying east at 250 mi/h and airplane B is flying
north at 300 mi/h. If they are both heading to the same
airport, located 30 miles east of airplane A and 40 miles
north of airplane B, at what rate is the distance between the
airplanes changing?
dt
Find dz at (x, y) = (1, 3) and z 2 = x 2 + y 2 if
dt
dx = 4 and dy = 3.
dt
dt
For the following exercises, sketch the situation if
necessary and used related rates to solve for the quantities.
4. [T] If two electrical resistors are connected in parallel,
the total resistance (measured in ohms, denoted by the
Greek capital letter omega, Ω) is given by the equation
1 = 1 + 1 . If R is increasing at a rate of 0.5 Ω/min
1
R R1 R2
and R 2 decreases at a rate of 1.1Ω/min, at what rate
does the total resistance change when R 1 = 20Ω and
R 2 = 50Ω/min ?
5. A 10-ft ladder is leaning against a wall. If the top of the
ladder slides down the wall at a rate of 2 ft/sec, how fast
is the bottom moving along the ground when the bottom of
the ladder is 5 ft from the wall?
8. You and a friend are riding your bikes to a restaurant
that you think is east; your friend thinks the restaurant is
north. You both leave from the same point, with you riding
at 16 mph east and your friend riding 12 mph north. After
you traveled 4 mi, at what rate is the distance between you
changing?
9. Two buses are driving along parallel freeways that are
5 mi apart, one heading east and the other heading west.
Assuming that each bus drives a constant 55 mph, find the
rate at which the distance between the buses is changing
when they are 13 mi apart, heading toward each other.
10. A 6-ft-tall person walks away from a 10-ft lamppost at
a constant rate of 3 ft/sec. What is the rate that the tip of
the shadow moves away from the pole when the person is
10 ft away from the pole?
6. A 25-ft ladder is leaning against a wall. If we push the
ladder toward the wall at a rate of 1 ft/sec, and the bottom
of the ladder is initially 20 ft away from the wall, how
fast does the ladder move up the wall 5 sec after we start
pushing?
11. Using the previous problem, what is the rate at which
the tip of the shadow moves away from the person when the
person is 10 ft from the pole?
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Chapter 4 | Applications of Derivatives
351
12. A 5-ft-tall person walks toward a wall at a rate of 2
ft/sec. A spotlight is located on the ground 40 ft from the
wall. How fast does the height of the person’s shadow on
the wall change when the person is 10 ft from the wall?
24. A triangle has a height that is increasing at a rate of 2
cm/sec and its area is increasing at a rate of 4 cm2/sec. Find
the rate at which the base of the triangle is changing when
the height of the triangle is 4 cm and the area is 20 cm2.
13. Using the previous problem, what is the rate at which
the shadow changes when the person is 10 ft from the wall,
if the person is walking away from the wall at a rate of 2 ft/
sec?
For the following exercises, consider a right cone that is
leaking water. The dimensions of the conical tank are a
height of 16 ft and a radius of 5 ft.
14. A helicopter starting on the ground is rising directly
into the air at a rate of 25 ft/sec. You are running on the
ground starting directly under the helicopter at a rate of 10
ft/sec. Find the rate of change of the distance between the
helicopter and yourself after 5 sec.
15. Using the previous problem, what is the rate at which
the distance between you and the helicopter is changing
when the helicopter has risen to a height of 60 ft in the air,
assuming that, initially, it was 30 ft above you?
For the following exercises, draw and label diagrams to
help solve the related-rates problems.
16. The side of a cube increases at a rate of 1 m/sec. Find
2
the rate at which the volume of the cube increases when the
side of the cube is 4 m.
17. The volume of a cube decreases at a rate of 10 m/sec.
Find the rate at which the side of the cube changes when
the side of the cube is 2 m.
18. The radius of a circle increases at a rate of 2 m/sec.
Find the rate at which the area of the circle increases when
the radius is 5 m.
19. The radius of a sphere decreases at a rate of 3 m/sec.
Find the rate at which the surface area decreases when the
radius is 10 m.
20. The radius of a sphere increases at a rate of 1 m/sec.
Find the rate at which the volume increases when the radius
is 20 m.
21. The radius of a sphere is increasing at a rate of 9 cm/
sec. Find the radius of the sphere when the volume and the
radius of the sphere are increasing at the same numerical
rate.
22. The base of a triangle is shrinking at a rate of 1 cm/min
and the height of the triangle is increasing at a rate of 5 cm/
min. Find the rate at which the area of the triangle changes
when the height is 22 cm and the base is 10 cm.
23. A triangle has two constant sides of length 3 ft and 5
ft. The angle between these two sides is increasing at a rate
of 0.1 rad/sec. Find the rate at which the area of the triangle
is changing when the angle between the two sides is π/6.
25. How fast does the depth of the water change when the
water is 10 ft high if the cone leaks water at a rate of 10
ft3/min?
26. Find the rate at which the surface area of the water
changes when the water is 10 ft high if the cone leaks water
at a rate of 10 ft3/min.
27. If the water level is decreasing at a rate of 3 in./min
when the depth of the water is 8 ft, determine the rate at
which water is leaking out of the cone.
28. A vertical cylinder is leaking water at a rate of 1
ft3/sec. If the cylinder has a height of 10 ft and a radius of 1
ft, at what rate is the height of the water changing when the
height is 6 ft?
29. A cylinder is leaking water but you are unable to
determine at what rate. The cylinder has a height of 2 m
and a radius of 2 m. Find the rate at which the water is
leaking out of the cylinder if the rate at which the height is
decreasing is 10 cm/min when the height is 1 m.
30. A trough has ends shaped like isosceles triangles,
with width 3 m and height 4 m, and the trough is 10
m long. Water is being pumped into the trough at a rate
of 5 m 3 /min. At what rate does the height of the water
change when the water is 1 m deep?
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Chapter 4 | Applications of Derivatives
31. A tank is shaped like an upside-down square pyramid,
with base of 4 m by 4 m and a height of 12 m (see the
following figure). How fast does the height increase when
the water is 2 m deep if water is being pumped in at a rate
of 2 m/sec?
3
39. A lighthouse, L, is on an island 4 mi away from the
closest point, P, on the beach (see the following image). If
the lighthouse light rotates clockwise at a constant rate of
10 revolutions/min, how fast does the beam of light move
across the beach 2 mi away from the closest point on the
beach?
For the following problems, consider a pool shaped like the
bottom half of a sphere, that is being filled at a rate of 25
ft3/min. The radius of the pool is 10 ft.
32. Find the rate at which the depth of the water is
changing when the water has a depth of 5 ft.
33. Find the rate at which the depth of the water is
changing when the water has a depth of 1 ft.
34. If the height is increasing at a rate of 1 in./sec when
the depth of the water is 2 ft, find the rate at which water is
being pumped in.
35. Gravel is being unloaded from a truck and falls into a
pile shaped like a cone at a rate of 10 ft3/min. The radius of
the cone base is three times the height of the cone. Find the
rate at which the height of the gravel changes when the pile
has a height of 5 ft.
36. Using a similar setup from the preceding problem, find
the rate at which the gravel is being unloaded if the pile is
5 ft high and the height is increasing at a rate of 4 in./min.
For the following exercises, draw the situations and solve
the related-rate problems.
37. You are stationary on the ground and are watching
a bird fly horizontally at a rate of 10 m/sec. The bird is
located 40 m above your head. How fast does the angle of
elevation change when the horizontal distance between you
and the bird is 9 m?
38. You stand 40 ft from a bottle rocket on the ground and
watch as it takes off vertically into the air at a rate of 20 ft/
sec. Find the rate at which the angle of elevation changes
when the rocket is 30 ft in the air.
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40. Using the same setup as the previous problem,
determine at what rate the beam of light moves across the
beach 1 mi away from the closest point on the beach.
41. You are walking to a bus stop at a right-angle corner.
You move north at a rate of 2 m/sec and are 20 m south
of the intersection. The bus travels west at a rate of 10 m/
sec away from the intersection – you have missed the bus!
What is the rate at which the angle between you and the bus
is changing when you are 20 m south of the intersection and
the bus is 10 m west of the intersection?
For the following exercises, refer to the figure of baseball
diamond, which has sides of 90 ft.
Chapter 4 | Applications of Derivatives
42. [T] A batter hits a ball toward third base at 75 ft/sec
and runs toward first base at a rate of 24 ft/sec. At what rate
does the distance between the ball and the batter change
when 2 sec have passed?
43. [T] A batter hits a ball toward second base at 80 ft/sec
and runs toward first base at a rate of 30 ft/sec. At what rate
does the distance between the ball and the batter change
when the runner has covered one-third of the distance to
first base? (Hint: Recall the law of cosines.)
44. [T] A batter hits the ball and runs toward first base at
a speed of 22 ft/sec. At what rate does the distance between
the runner and second base change when the runner has run
30 ft?
45. [T] Runners start at first and second base. When the
baseball is hit, the runner at first base runs at a speed of
18 ft/sec toward second base and the runner at second base
runs at a speed of 20 ft/sec toward third base. How fast is
the distance between runners changing 1 sec after the ball
is hit?
353
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Chapter 4 | Applications of Derivatives
4.2 | Linear Approximations and Differentials
Learning Objectives
4.2.1 Describe the linear approximation to a function at a point.
4.2.2 Write the linearization of a given function.
4.2.3 Draw a graph that illustrates the use of differentials to approximate the change in a
quantity.
4.2.4 Calculate the relative error and percentage error in using a differential approximation.
We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, we
examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions
are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition,
the ideas presented in this section are generalized later in the text when we study how to approximate functions by higherdegree polynomials Introduction to Power Series and Functions (http://cnx.org/content/m53760/latest/) .
Linear Approximation of a Function at a Point
Consider a function f that is differentiable at a point x = a. Recall that the tangent line to the graph of f at a is given
by the equation
y = f (a) + f ′(a)(x − a).
For example, consider the function f (x) = 1x at a = 2. Since f is differentiable at x = 2 and f ′(x) = − 12 , we see
x
that f ′(2) = − 1 . Therefore, the tangent line to the graph of f at a = 2 is given by the equation
4
y = 1 − 1 (x − 2).
2 4
Figure 4.7(a) shows a graph of f (x) = 1x along with the tangent line to f at x = 2. Note that for x near 2, the graph of
the tangent line is close to the graph of f . As a result, we can use the equation of the tangent line to approximate f (x) for
x near 2. For example, if x = 2.1, the y value of the corresponding point on the tangent line is
y = 1 − 1 (2.1 − 2) = 0.475.
2 4
The actual value of f (2.1) is given by
f (2.1) = 1 ≈ 0.47619.
2.1
Therefore, the tangent line gives us a fairly good approximation of f (2.1) (Figure 4.7(b)). However, note that for values
of x far from 2, the equation of the tangent line does not give us a good approximation. For example, if x = 10, the y
-value of the corresponding point on the tangent line is
y = 1 − 1 (10 − 2) = 1 − 2 = −1.5,
2
2 4
whereas the value of the function at x = 10 is f (10) = 0.1.
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Chapter 4 | Applications of Derivatives
355
Figure 4.7 (a) The tangent line to f (x) = 1/x at x = 2 provides a good approximation to f for x near 2.
(b) At x = 2.1, the value of y on the tangent line to f (x) = 1/x is 0.475. The actual value of f (2.1) is
1/2.1, which is approximately 0.47619.
In general, for a differentiable function f , the equation of the tangent line to f at x = a can be used to approximate
f (x) for x near a. Therefore, we can write
f (x) ≈ f (a) + f ′(a)(x − a) for x near a.
We call the linear function
L(x) = f (a) + f ′(a)(x − a)
(4.1)
the linear approximation, or tangent line approximation, of f at x = a. This function L is also known as the
linearization of f at x = a.
To show how useful the linear approximation can be, we look at how to find the linear approximation for f (x) = x at
x = 9.
Example 4.5
Linear Approximation of x
Find the linear approximation of f (x) = x at x = 9 and use the approximation to estimate
9.1.
Solution
Since we are looking for the linear approximation at x = 9,
using Equation 4.1 we know the linear
approximation is given by
L(x) = f (9) + f ′(9)(x − 9).
We need to find f (9) and f ′(9).
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Chapter 4 | Applications of Derivatives
f (x) = x ⇒ f (9) = 9 = 3
f ′(x) = 1 ⇒ f ′(9) = 1 = 1
2 x
2 9 6
Therefore, the linear approximation is given by Figure 4.8.
L(x) = 3 + 1 (x − 9)
6
Using the linear approximation, we can estimate
9.1 by writing
9.1 = f (9.1) ≈ L(9.1) = 3 + 1 (9.1 − 9) ≈ 3.0167.
6
Figure 4.8 The local linear approximation to f (x) = x at
x = 9 provides an approximation to f for x near 9.
Analysis
Using a calculator, the value of 9.1 to four decimal places is 3.0166. The value given by the linear
approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear
approximation is a good way to estimate x, at least for x near 9. At the same time, it may seem odd to use
a linear approximation when we can just push a few buttons on a calculator to evaluate
9.1. However, how
does the calculator evaluate 9.1 ? The calculator uses an approximation! In fact, calculators and computers use
approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.
4.5
3
Find the local linear approximation to f (x) = 3 x at x = 8. Use it to approximate 8.1 to five decimal
places.
Example 4.6
Linear Approximation of sinx
Find the linear approximation of f (x) = sin x at x = π and use it to approximate sin(62°).
3
Solution
First we note that since π rad is equivalent to 60°,
3
reasonable. The linear approximation is given by
using the linear approximation at x = π/3 seems
⎛ ⎞
⎛ ⎞⎛
⎞
L(x) = f ⎝π ⎠ + f ′⎝π ⎠⎝x − π ⎠.
3
3
3
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Chapter 4 | Applications of Derivatives
357
We see that
⎛ ⎞
⎛ ⎞
f (x) = sin x ⇒ f ⎝π ⎠ = sin⎝π ⎠ = 3
3
3
2
⎛π ⎞
⎛π ⎞ 1
f ′(x) = cos x ⇒ f ′⎝ ⎠ = cos⎝ ⎠ =
3
3
2
Therefore, the linear approximation of f at x = π/3 is given by Figure 4.9.
⎛
⎞
L(x) = 3 + 1 ⎝x − π ⎠
3
2
2
To estimate sin(62°) using L, we must first convert 62° to radians. We have 62° = 62π radians, so the
estimate for sin(62°) is given by
180
⎛
⎞
⎛
⎞
⎛
⎞
⎛
⎞
sin(62°) = f ⎝62π ⎠ ≈ L⎝62π ⎠ = 3 + 1 ⎝62π − π ⎠ = 3 + 1 ⎝ 2π ⎠ = 3 + π ≈ 0.88348.
180
180
180
2
2 180 3
2
2 180
2
Figure 4.9 The linear approximation to f (x) = sin x at x = π/3 provides an approximation
to sin x for x near π/3.
4.6
Find the linear approximation for f (x) = cos x at x = π .
2
Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation
for f (x) = (1 + x) n at x = 0, which can be used to estimate roots and powers for real numbers near 1. The same idea
can be extended to a function of the form f (x) = (m + x) n to estimate roots and powers near a different number m.
Example 4.7
Approximating Roots and Powers
Find the linear approximation of f (x) = (1 + x) n at x = 0. Use this approximation to estimate (1.01) 3.
Solution
The linear approximation at x = 0 is given by
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Chapter 4 | Applications of Derivatives
L(x) = f (0) + f ′(0)(x − 0).
Because
f (x) = (1 + x) n ⇒ f (0) = 1
f ′(x) = n(1 + x) n − 1 ⇒ f ′(0) = n,
the linear approximation is given by Figure 4.10(a).
L(x) = 1 + n(x − 0) = 1 + nx
We can approximate (1.01) 3 by evaluating L(0.01) when n = 3. We conclude that
(1.01) 3 = f (1.01) ≈ L(1.01) = 1 + 3(0.01) = 1.03.
Figure 4.10 (a) The linear approximation of f (x) at x = 0 is L(x). (b) The actual value of 1.01 3 is
1.030301. The linear approximation of f (x) at x = 0 estimates 1.01 3 to be 1.03.
4.7
Find the linear approximation of f (x) = (1 + x) 4 at x = 0 without using the result from the preceding
example.
Differentials
We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the
amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related
concept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a small
change in input values.
When we first looked at derivatives, we used the Leibniz notation dy/dx to represent the derivative of y with respect to
x. Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a
meaning to the expressions dy and dx. Suppose y = f (x) is a differentiable function. Let dx be an independent variable that
can be assigned any nonzero real number, and define the dependent variable dy by
dy = f ′(x)dx.
(4.2)
It is important to notice that dy is a function of both x and dx. The expressions dy and dx are called differentials. We can
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Chapter 4 | Applications of Derivatives
359
divide both sides of Equation 4.2 by dx, which yields
dy
= f ′(x).
dx
(4.3)
This is the familiar expression we have used to denote a derivative. Equation 4.2 is known as the differential form of
Equation 4.3.
Example 4.8
Computing differentials
For each of the following functions, find dy and evaluate when x = 3 and dx = 0.1.
a.
y = x 2 + 2x
b.
y = cos x
Solution
The key step is calculating the derivative. When we have that, we can obtain dy directly.
a. Since f (x) = x 2 + 2x, we know f ′(x) = 2x + 2, and therefore
dy = (2x + 2)dx.
When x = 3 and dx = 0.1,
dy = (2 · 3 + 2)(0.1) = 0.8.
b. Since f (x) = cos x,
f ′(x) = −sin(x). This gives us
dy = −sin x dx.
When x = 3 and dx = 0.1,
dy = −sin(3)(0.1) = −0.1sin(3).
4.8
2
For y = e x , find dy.
We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a
function resulting from a small change in input values. Consider a function f that is differentiable at point a. Suppose
the input x changes by a small amount. We are interested in how much the output y changes. If x changes from a to
a + dx, then the change in x is dx (also denoted Δx), and the change in y is given by
Δy = f (a + dx) − f (a).
Instead of calculating the exact change in y, however, it is often easier to approximate the change in y by using a linear
approximation. For x near a,
f (x) can be approximated by the linear approximation
L(x) = f (a) + f ′(a)(x − a).
Therefore, if dx is small,
f (a + dx) ≈ L(a + dx) = f (a) + f ′(a)(a + dx − a).
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Chapter 4 | Applications of Derivatives
That is,
f (a + dx) − f (a) ≈ L(a + dx) − f (a) = f ′(a)dx.
In other words, the actual change in the function f if x increases from a to a + dx is approximately the difference
between L(a + dx) and f (a), where L(x) is the linear approximation of f at a. By definition of L(x), this difference
is equal to f ′(a)dx. In summary,
Δy = f (a + dx) − f (a) ≈ L(a + dx) − f (a) = f ′(a)dx = dy.
Therefore, we can use the differential dy = f ′(a)dx to approximate the change in y if x increases from x = a to
x = a + dx. We can see this in the following graph.
Figure 4.11 The differential dy = f ′(a)dx is used to approximate the actual
change in y if x increases from a to a + dx.
We now take a look at how to use differentials to approximate the change in the value of the function that results from a
small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values
of functions and the result is very close to what we would obtain with the more exact calculation.
Example 4.9
Approximating Change with Differentials
Let y = x 2 + 2x. Compute Δy and dy at x = 3 if dx = 0.1.
Solution
The actual change in y if x changes from x = 3 to x = 3.1 is given by
Δy = f (3.1) − f (3) = [(3.1) 2 + 2(3.1)] − [3 2 + 2(3)] = 0.81.
The approximate change in y is given by dy = f ′(3)dx. Since f ′(x) = 2x + 2, we have
dy = f ′(3)dx = (2(3) + 2)(0.1) = 0.8.
4.9
For y = x 2 + 2x, find Δy and dy at x = 3 if dx = 0.2.
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Chapter 4 | Applications of Derivatives
361
Calculating the Amount of Error
Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based
on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the
measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study
how differentials can be used to estimate the error.
Consider a function f with an input that is a measured quantity. Suppose the exact value of the measured quantity is a,
but the measured value is a + dx. We say the measurement error is dx (or Δx). As a result, an error occurs in the calculated
quantity f (x). This type of error is known as a propagated error and is given by
Δy = f (a + dx) − f (a).
Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we
cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use
differentials to approximate the propagated error Δy. Specifically, if f is a differentiable function at a, the propagated
error is
Δy ≈ dy = f ′(a)dx.
Unfortunately, we do not know the exact value a. However, we can use the measured value a + dx, and estimate
Δy ≈ dy ≈ f ′(a + dx)dx.
In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we
assume the measurement of the side length is made with a certain amount of accuracy.
Example 4.10
Volume of a Cube
Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.
a. Use differentials to estimate the error in the computed volume of the cube.
b. Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated
error with the actual potential error.
Solution
a. The measurement of the side length is accurate to within ±0.1 cm. Therefore,
−0.1 ≤ dx ≤ 0.1.
The volume of a cube is given by V = x 3, which leads to
dV = 3x 2 dx.
Using the measured side length of 5 cm, we can estimate that
−3(5) 2(0.1) ≤ dV ≤ 3(5) 2(0.1).
Therefore,
−7.5 ≤ dV ≤ 7.5.
b. If the side length is actually 4.9 cm, then the volume of the cube is
V(4.9) = (4.9) 3 = 117.649 cm 3.
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Chapter 4 | Applications of Derivatives
If the side length is actually 5.1 cm, then the volume of the cube is
V(5.1) = (5.1) 3 = 132.651 cm 3.
Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is
measured to be 5 cm, the computed volume is V(5) = 5 3 = 125. Therefore, the error in the computed
volume is
117.649 − 125 ≤ ΔV ≤ 132.651 − 125.
That is,
−7.351 ≤ ΔV ≤ 7.651.
We see the estimated error dV is relatively close to the actual potential error in the computed volume.
4.10 Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an
accuracy of 0.2 cm.
The measurement error dx (=Δx) and the propagated error Δy are absolute errors. We are typically interested in the size
of an error relative to the size of the quantity being measured or calculated. Given an absolute error Δq for a particular
Δq
quantity, we define the relative error as q , where q is the actual value of the quantity. The percentage error is the
relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual
height is 62 in., the absolute error is 1 in. but the relative error is 1 = 0.016, or 1.6%. By comparison, if we measure the
62
width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is 1 in., whereas the relative
4
error is 0.25 = 1 , or 3.1%. Therefore, the percentage error in the measurement of the cardboard is larger, even though
8
32
0.25 in. is less than 1 in.
Example 4.11
Relative and Percentage Error
An astronaut using a camera measures the radius of Earth as 4000 mi with an error of ±80 mi. Let’s use
differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume
of Earth, assuming the planet is a perfect sphere.
Solution
If the measurement of the radius is accurate to within ±80, we have
−80 ≤ dr ≤ 80.
⎛ ⎞
Since the volume of a sphere is given by V = ⎝4 ⎠πr 3, we have
3
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Chapter 4 | Applications of Derivatives
363
dV = 4πr 2 dr.
Using the measured radius of 4000 mi, we can estimate
−4π(4000) 2(80) ≤ dV ≤ 4π(4000) 2(80).
To estimate the relative error, consider dV . Since we do not know the exact value of the volume V, use the
V
⎛ ⎞
measured radius r = 4000 mi to estimate V. We obtain V ≈ ⎝4 ⎠π(4000) 3. Therefore the relative error satisfies
3
−4π(4000) 2(80) dV 4π(4000) 2(80)
≤
≤
,
V
4π(4000) 3 /3
4π(4000) 3 /3
which simplifies to
−0.06 ≤ dV ≤ 0.06.
V
The relative error is 0.06 and the percentage error is 6%.
4.11
mi.
Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of ±100
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Chapter 4 | Applications of Derivatives
4.2 EXERCISES
46. What is the linear approximation for any generic linear
function y = mx + b ?
47. Determine the necessary conditions such that the
linear approximation function is constant. Use a graph to
prove your result.
48. Explain why the linear approximation becomes less
accurate as you increase the distance between x and a.
Use a graph to prove your argument.
49. When is the linear approximation exact?
For the following exercises, find the linear approximation
L(x) to y = f (x) near x = a for the function.
50. [T] f (x) = x + x 4, a = 0
51. [T] f (x) = 1x , a = 2
52. [T] f (x) = tan x, a = π
4
53. [T] f (x) = sin x, a = π
2
54. [T] f (x) = x sin x, a = 2π
55. [T] f (x) = sin 2 x, a = 0
62. (1.01) 3
63. cos(0.01)
64.
sin(0.01)⎞⎠ 2
⎛
⎝
65. (1.01) −3
⎛
⎞
66. ⎝1 + 1 ⎠
10
67.
10
8.99
For the following exercises, find the differential of the
function.
68. y = 3x 4 + x 2 − 2x + 1
69. y = x cos x
70. y = 1 + x
2
71. y = x + 2
x−1
For the following exercises, find the differential and
evaluate for the given x and dx.
For the following exercises, compute the values given
within 0.01 by deciding on the appropriate f (x) and a,
72. y = 3x 2 − x + 6,
and evaluating L(x) = f (a) + f ′(a)(x − a). Check your
73. y =
answer using a calculator.
56. [T] (2.001) 6
57. [T] sin(0.02)
58. [T] cos(0.03)
59. [T] (15.99) 1/4
60. [T] 1
0.98
61. [T] sin(3.14)
For the following exercises, determine the appropriate
f (x) and a, and evaluate L(x) = f (a) + f ′ (a)(x − a).
Calculate the numerical error in the linear approximations
that follow.
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1 , x = 1, dx = 0.25
x+1
74. y = tan x,
x = 0, dx = π
10
2
75. y = 3x + 2 ,
x+1
76. y =
x = 2, dx = 0.1
x = 0, dx = 0.1
sin(2x)
x , x = π, dx = 0.25
77. y = x 3 + 2x + 1x ,
x = 1, dx = 0.05
For the following exercises, find the change in volume dV
or in surface area dA.
78. dV if the sides of a cube change from 10 to 10.1.
79. dA if the sides of a cube change from x to x + dx.
80. dA if the radius of a sphere changes from r by dr.
Chapter 4 | Applications of Derivatives
81. dV if the radius of a sphere changes from r by dr.
82. dV if a circular cylinder with r = 2 changes height
from 3 cm to 3.05 cm.
dV if a circular cylinder of height 3 changes from
r = 2 to r = 1.9 cm.
83.
For the following exercises, use differentials to estimate the
maximum and relative error when computing the surface
area or volume.
84. A spherical golf ball is measured to have a radius of
5 mm, with a possible measurement error of 0.1 mm.
What is the possible change in volume?
85. A pool has a rectangular base of 10 ft by 20 ft and a
depth of 6 ft. What is the change in volume if you only fill
it up to 5.5 ft?
86. An ice cream cone has height 4 in. and radius 1 in. If
the cone is 0.1 in. thick, what is the difference between the
volume of the cone, including the shell, and the volume of
the ice cream you can fit inside the shell?
For the following exercises, confirm the approximations by
using the linear approximation at x = 0.
87.
1 − x ≈ 1 − 1x
2
88.
1
≈1
1 − x2
89.
c2 + x2 ≈ c
365
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Chapter 4 | Applications of Derivatives
4.3 | Maxima and Minima
Learning Objectives
4.3.1
4.3.2
4.3.3
4.3.4
Define absolute extrema.
Define local extrema.
Explain how to find the critical points of a function over a closed interval.
Describe how to use critical points to locate absolute extrema over a closed interval.
Given a particular function, we are often interested in determining the largest and smallest values of the function. This
information is important in creating accurate graphs. Finding the maximum and minimum values of a function also
has practical significance because we can use this method to solve optimization problems, such as maximizing profit,
minimizing the amount of material used in manufacturing an aluminum can, or finding the maximum height a rocket can
reach. In this section, we look at how to use derivatives to find the largest and smallest values for a function.
Absolute Extrema
Consider the function f (x) = x 2 + 1 over the interval (−∞, ∞). As x → ±∞,
f (x) → ∞. Therefore, the function
2
does not have a largest value. However, since x + 1 ≥ 1 for all real numbers x and x 2 + 1 = 1 when x = 0, the
function has a smallest value, 1, when x = 0. We say that 1 is the absolute minimum of f (x) = x 2 + 1 and it occurs at
x = 0. We say that f (x) = x 2 + 1 does not have an absolute maximum (see the following figure).
Figure 4.12 The given function has an absolute minimum of 1
at x = 0. The function does not have an absolute maximum.
Definition
Let f be a function defined over an interval I and let c ∈ I. We say f has an absolute maximum on I at c if
f (c) ≥ f (x) for all x ∈ I. We say f has an absolute minimum on I at c if f (c) ≤ f (x) for all x ∈ I. If f has
an absolute maximum on I at c or an absolute minimum on I at c, we say f has an absolute extremum on I at
c.
Before proceeding, let’s note two important issues regarding this definition. First, the term absolute here does not refer to
absolute value. An absolute extremum may be positive, negative, or zero. Second, if a function f has an absolute extremum
over an interval I at c, the absolute extremum is f (c). The real number c is a point in the domain at which the absolute
extremum occurs. For example, consider the function f (x) = 1/(x 2 + 1) over the interval (−∞, ∞). Since
f (0) = 1 ≥
1 = f (x)
x2 + 1
for all real numbers x, we say f has an absolute maximum over (−∞, ∞) at x = 0. The absolute maximum is
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Chapter 4 | Applications of Derivatives
367
f (0) = 1. It occurs at x = 0, as shown in Figure 4.13(b).
A function may have both an absolute maximum and an absolute minimum, just one extremum, or neither. Figure 4.13
shows several functions and some of the different possibilities regarding absolute extrema. However, the following theorem,
called the Extreme Value Theorem, guarantees that a continuous function f over a closed, bounded interval [a, b] has
both an absolute maximum and an absolute minimum.
Figure 4.13 Graphs (a), (b), and (c) show several possibilities for absolute extrema for functions with a domain of
(−∞, ∞). Graphs (d), (e), and (f) show several possibilities for absolute extrema for functions with a domain that is a
bounded interval.
Theorem 4.1: Extreme Value Theorem
If f is a continuous function over the closed, bounded interval [a, b], then there is a point in [a, b] at which f has
an absolute maximum over [a, b] and there is a point in [a, b] at which f has an absolute minimum over [a, b].
The proof of the extreme value theorem is beyond the scope of this text. Typically, it is proved in a course on real analysis.
There are a couple of key points to note about the statement of this theorem. For the extreme value theorem to apply, the
368
Chapter 4 | Applications of Derivatives
function must be continuous over a closed, bounded interval. If the interval I is open or the function has even one point
of discontinuity, the function may not have an absolute maximum or absolute minimum over I. For example, consider the
functions shown in Figure 4.13(d), (e), and (f). All three of these functions are defined over bounded intervals. However,
the function in graph (e) is the only one that has both an absolute maximum and an absolute minimum over its domain.
The extreme value theorem cannot be applied to the functions in graphs (d) and (f) because neither of these functions is
continuous over a closed, bounded interval. Although the function in graph (d) is defined over the closed interval [0, 4],
the function is discontinuous at x = 2. The function has an absolute maximum over [0, 4] but does not have an absolute
minimum. The function in graph (f) is continuous over the half-open interval [0, 2), but is not defined at x = 2, and
therefore is not continuous over a closed, bounded interval. The function has an absolute minimum over [0, 2), but does
not have an absolute maximum over [0, 2). These two graphs illustrate why a function over a bounded interval may fail to
have an absolute maximum and/or absolute minimum.
Before looking at how to find absolute extrema, let’s examine the related concept of local extrema. This idea is useful in
determining where absolute extrema occur.
Local Extrema and Critical Points
Consider the function f shown in Figure 4.14. The graph can be described as two mountains with a valley in the middle.
The absolute maximum value of the function occurs at the higher peak, at x = 2. However, x = 0 is also a point of
interest. Although f (0) is not the largest value of f , the value f (0) is larger than f (x) for all x near 0. We say f has a
local maximum at x = 0. Similarly, the function f does not have an absolute minimum, but it does have a local minimum
at x = 1 because f (1) is less than f (x) for x near 1.
Figure 4.14 This function f has two local maxima and one
local minimum. The local maximum at x = 2 is also the
absolute maximum.
Definition
A function f has a local maximum at c if there exists an open interval I containing c such that I is contained
in the domain of f and f (c) ≥ f (x) for all x ∈ I. A function f has a local minimum at c if there exists an open
interval I containing c such that I is contained in the domain of f and f (c) ≤ f (x) for all x ∈ I. A function f
has a local extremum at c if f has a local maximum at c or f has a local minimum at c.
Note that if f has an absolute extremum at c and f is defined over an interval containing c, then f (c) is also
considered a local extremum. If an absolute extremum for a function f occurs at an endpoint, we do not consider that to be
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Chapter 4 | Applications of Derivatives
369
a local extremum, but instead refer to that as an endpoint extremum.
Given the graph of a function f , it is sometimes easy to see where a local maximum or local minimum occurs. However,
it is not always easy to see, since the interesting features on the graph of a function may not be visible because they occur at
a very small scale. Also, we may not have a graph of the function. In these cases, how can we use a formula for a function
to determine where these extrema occur?
To answer this question, let’s look at Figure 4.14 again. The local extrema occur at x = 0,
x = 1, and x = 2. Notice
that at x = 0 and x = 1, the derivative f ′(x) = 0. At x = 2, the derivative f ′(x) does not exist, since the function
f has a corner there. In fact, if f has a local extremum at a point x = c, the derivative f ′(c) must satisfy one of the
following conditions: either f ′(c) = 0 or f ′(c) is undefined. Such a value c is known as a critical point and it is important
in finding extreme values for functions.
Definition
Let c be an interior point in the domain of f . We say that c is a critical point of f if f ′(c) = 0 or f ′(c) is
undefined.
As mentioned earlier, if f has a local extremum at a point x = c, then c must be a critical point of f . This fact is known
as Fermat’s theorem.
Theorem 4.2: Fermat’s Theorem
If f has a local extremum at c and f is differentiable at c, then f ′(c) = 0.
Proof
Suppose f has a local extremum at c and f is differentiable at c. We need to show that f ′(c) = 0. To do this, we
will show that f ′(c) ≥ 0 and f ′(c) ≤ 0, and therefore f ′(c) = 0. Since f has a local extremum at c,
f has a local
maximum or local minimum at c. Suppose f has a local maximum at c. The case in which f has a local minimum
at c can be handled similarly. There then exists an open interval I such that f (c) ≥ f (x) for all x ∈ I. Since f is
differentiable at c, from the definition of the derivative, we know that
f ′(c) = xlim
→c
f (x) − f (c)
x−c .
Since this limit exists, both one-sided limits also exist and equal f ′(c). Therefore,
f ′(c) = lim
f (x) − f (c)
x−c ,
(4.4)
f ′(c) = lim−
f (x) − f (c)
x−c .
(4.5)
x → c+
and
x→c
Since f (c) is a local maximum, we see that f (x) − f (c) ≤ 0 for x near c. Therefore, for x near c, but x > c,
f (x) − f (c)
≤ 0. From Equation 4.4 we conclude that f ′(c) ≤ 0. Similarly, it can be shown that f ′(c) ≥ 0.
x−c
Therefore, f ′(c) = 0.
we have
□
From Fermat’s theorem, we conclude that if f has a local extremum at c, then either f ′(c) = 0 or f ′(c) is undefined.
In other words, local extrema can only occur at critical points.
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Chapter 4 | Applications of Derivatives
Note this theorem does not claim that a function f must have a local extremum at a critical point. Rather, it states that
critical points are candidates for local extrema. For example, consider the function f (x) = x 3. We have f ′(x) = 3x 2 = 0
when x = 0. Therefore, x = 0 is a critical point. However, f (x) = x 3 is increasing over (−∞, ∞), and thus f does
not have a local extremum at x = 0. In Figure 4.15, we see several different possibilities for critical points. In some of
these cases, the functions have local extrema at critical points, whereas in other cases the functions do not. Note that these
graphs do not show all possibilities for the behavior of a function at a critical point.
Figure 4.15 (a–e) A function f has a critical point at c if f ′(c) = 0 or f ′(c) is undefined. A function may or may not
have a local extremum at a critical point.
Later in this chapter we look at analytical methods for determining whether a function actually has a local extremum at a
critical point. For now, let’s turn our attention to finding critical points. We will use graphical observations to determine
whether a critical point is associated with a local extremum.
Example 4.12
Locating Critical Points
For each of the following functions, find all critical points. Use a graphing utility to determine whether the
function has a local extremum at each of the critical points.
a.
f (x) = 1 x 3 − 5 x 2 + 4x
3
2
b.
f (x) = ⎛⎝x 2 − 1⎞⎠
c.
f (x) =
3
4x
1 + x2
Solution
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Chapter 4 | Applications of Derivatives
371
a. The derivative f ′(x) = x 2 − 5x + 4 is defined for all real numbers x. Therefore, we only need to find
the values for x where f ′(x) = 0. Since f ′(x) = x 2 − 5x + 4 = (x − 4)(x − 1), the critical points are
x = 1 and x = 4. From the graph of f in Figure 4.16, we see that f has a local maximum at x = 1
and a local minimum at x = 4.
Figure 4.16 This function has a local maximum and a local
minimum.
b. Using the chain rule, we see the derivative is
2
2
f ′(x) = 3⎛⎝x 2 − 1⎞⎠ (2x) = 6x⎛⎝x 2 − 1⎞⎠ .
Therefore, f has critical points when x = 0 and when x 2 − 1 = 0. We conclude that the critical points
are x = 0, ±1. From the graph of f in Figure 4.17, we see that f has a local (and absolute) minimum
at x = 0, but does not have a local extremum at x = 1 or x = −1.
Figure 4.17 This function has three critical points: x = 0,
x = 1, and x = −1. The function has a local (and absolute)
minimum at x = 0, but does not have extrema at the other two
critical points.
c. By the chain rule, we see that the derivative is
f ′(x) =
⎛
⎝1
+ x 2 4⎞⎠ − 4x(2x)
⎛
⎝1
2
⎞
+ x 2⎠
2
= 4 − 4x 2 .
⎛
2⎞
⎝1 + x ⎠
The derivative is defined everywhere. Therefore, we only need to find values for x where f ′(x) = 0.
Solving f ′(x) = 0, we see that 4 − 4x 2 = 0, which implies x = ±1. Therefore, the critical points
are x = ±1. From the graph of f in Figure 4.18, we see that f has an absolute maximum at x = 1
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Chapter 4 | Applications of Derivatives
and an absolute minimum at x = −1. Hence, f has a local maximum at x = 1 and a local minimum at
x = −1. (Note that if f has an absolute extremum over an interval I at a point c that is not an endpoint
of I, then f has a local extremum at c.)
Figure 4.18 This function has an absolute maximum and an
absolute minimum.
4.12
Find all critical points for f (x) = x 3 − 1 x 2 − 2x + 1.
2
Locating Absolute Extrema
The extreme value theorem states that a continuous function over a closed, bounded interval has an absolute maximum and
an absolute minimum. As shown in Figure 4.13, one or both of these absolute extrema could occur at an endpoint. If an
absolute extremum does not occur at an endpoint, however, it must occur at an interior point, in which case the absolute
extremum is a local extremum. Therefore, by Fermat’s Theorem, the point c at which the local extremum occurs must
be a critical point. We summarize this result in the following theorem.
Theorem 4.3: Location of Absolute Extrema
Let f be a continuous function over a closed, bounded interval I. The absolute maximum of f over I and the
absolute minimum of f over I must occur at endpoints of I or at critical points of f in I.
With this idea in mind, let’s examine a procedure for locating absolute extrema.
Problem-Solving Strategy: Locating Absolute Extrema over a Closed Interval
Consider a continuous function f defined over the closed interval [a, b].
1. Evaluate f at the endpoints x = a and x = b.
2. Find all critical points of f that lie over the interval (a, b) and evaluate f at those critical points.
3. Compare all values found in (1) and (2). From Location of Absolute Extrema, the absolute extrema must
occur at endpoints or critical points. Therefore, the largest of these values is the absolute maximum of f . The
smallest of these values is the absolute minimum of f .
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Chapter 4 | Applications of Derivatives
373
Now let’s look at how to use this strategy to find the absolute maximum and absolute minimum values for continuous
functions.
Example 4.13
Locating Absolute Extrema
For each of the following functions, find the absolute maximum and absolute minimum over the specified interval
and state where those values occur.
a.
f (x) = −x 2 + 3x − 2 over [1, 3].
b.
f (x) = x 2 − 3x 2/3 over [0, 2].
Solution
a. Step 1. Evaluate f at the endpoints x = 1 and x = 3.
f (1) = 0 and f (3) = −2
Step 2. Since f ′(x) = −2x + 3,
f ′ is defined for all real numbers x. Therefore, there are no critical
points where the derivative is undefined. It remains to check where
f ′(x) = −2x + 3 = 0 at x = 3 and 3 is in the interval [1, 3],
2
2
⎛ ⎞
extremum of f over [1, 3]. We evaluate f ⎝3 ⎠ and find
2
⎛ ⎞
f ′(x) = 0.
Since
f ⎝3 ⎠ is a candidate for an absolute
2
⎛ ⎞
f ⎝3 ⎠ = 1 .
4
2
Step 3. We set up the following table to compare the values found in steps 1 and 2.
x
f(x)
0
0
3
2
1
4
Absolute maximum
3
−2
Absolute minimum
Conclusion
From the table, we find that the absolute maximum of f over the interval [1, 3] is 1 , and it occurs at
4
3
x = . The absolute minimum of f over the interval [1, 3] is −2, and it occurs at x = 3 as shown in
2
the following graph.
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Chapter 4 | Applications of Derivatives
Figure 4.19 This function has both an absolute maximum and an absolute minimum.
b. Step 1. Evaluate f at the endpoints x = 0 and x = 2.
3
f (0) = 0 and f (2) = 4 − 3 4 ≈ − 0.762
Step 2. The derivative of f is given by
4/3
f ′(x) = 2x − 21/3 = 2x 1/3− 2
x
x
for x ≠ 0. The derivative is zero when 2x 4/3 − 2 = 0, which implies x = ±1. The derivative is
undefined at x = 0. Therefore, the critical points of f are x = 0, 1, −1. The point x = 0 is an
endpoint, so we already evaluated f (0) in step 1. The point x = −1 is not in the interval of interest, so
we need only evaluate f (1). We find that
f (1) = −2.
Step 3. We compare the values found in steps 1 and 2, in the following table.
x
f(x)
Conclusion
0
0
Absolute maximum
1
−2
Absolute minimum
2
−0.762
We conclude that the absolute maximum of f over the interval [0, 2] is zero, and it occurs at x = 0. The
absolute minimum is −2, and it occurs at x = 1 as shown in the following graph.
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Chapter 4 | Applications of Derivatives
375
Figure 4.20 This function has an absolute maximum at an
endpoint of the interval.
4.13
Find the absolute maximum and absolute minimum of f (x) = x 2 − 4x + 3 over the interval [1, 4].
At this point, we know how to locate absolute extrema for continuous functions over closed intervals. We have also defined
local extrema and determined that if a function f has a local extremum at a point c, then c must be a critical point of f .
However, c being a critical point is not a sufficient condition for f to have a local extremum at c. Later in this chapter,
we show how to determine whether a function actually has a local extremum at a critical point. First, however, we need to
introduce the Mean Value Theorem, which will help as we analyze the behavior of the graph of a function.
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Chapter 4 | Applications of Derivatives
4.3 EXERCISES
90. In precalculus, you learned a formula for the position
of the maximum or minimum of a quadratic equation
100.
y = ax 2 + bx + c, which was m = − b . Prove this
(2a)
formula using calculus.
91. If you are finding an absolute minimum over an
interval [a, b], why do you need to check the endpoints?
Draw a graph that supports your hypothesis.
92.
If you are examining a function over an interval
(a, b), for a and b finite, is it possible not to have an
absolute maximum or absolute minimum?
101.
93. When you are checking for critical points, explain why
you also need to determine points where f (x) is undefined.
Draw a graph to support your explanation.
94.
Can you have a finite absolute maximum for
y = ax 2 + bx + c over (−∞, ∞) ? Explain why or why
not using graphical arguments.
95.
Can you have a finite absolute maximum for
y = ax 3 + bx 2 + cx + d over (−∞, ∞) assuming a is
non-zero? Explain why or why not using graphical
arguments.
102.
96. Let m be the number of local minima and M be the
number of local maxima. Can you create a function where
M > m + 2 ? Draw a graph to support your explanation.
97. Is it possible to have more than one absolute
maximum? Use a graphical argument to prove your
hypothesis.
98. Is it possible to have no absolute minimum or
maximum for a function? If so, construct such a function.
If not, explain why this is not possible.
99. [T] Graph the function y = e ax. For which values
103.
of a, on any infinite domain, will you have an absolute
minimum and absolute maximum?
For the following exercises, determine where the local and
absolute maxima and minima occur on the graph given.
Assume domains are closed intervals unless otherwise
specified.
For the following problems, draw graphs of f (x), which
is continuous, over the interval [−4, 4] with the following
properties:
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Chapter 4 | Applications of Derivatives
377
104. Absolute maximum at x = 2 and absolute minima at
x = ±3
105. Absolute minimum at x = 1 and absolute maximum
at x = 2
106. Absolute maximum at x = 4, absolute minimum at
x = −1, local maximum at x = −2, and a critical point
that is not a maximum or minimum at x = 2
107.
Absolute maxima at x = 2 and x = −3,
local
minimum at x = 1, and absolute minimum at x = 4
123. y = x + sin(x) over [0, 2π]
124. y =
x over [0, 100]
1+x
125. y = |x + 1| + |x − 1| over [−3, 2]
126. y = x − x 3 over [0, 4]
127. y = sin x + cos x over [0, 2π]
128. y = 4sinθ − 3cosθ over [0, 2π]
For the following exercises, find the critical points in the
domains of the following functions.
For the following exercises, find the local and absolute
minima and maxima for the functions over (−∞, ∞).
108. y = 4x 3 − 3x
129. y = x 2 + 4x + 5
109. y = 4 x − x 2
130. y = x 3 − 12x
110. y =
131. y = 3x 4 + 8x 3 − 18x 2
1
x−1
111. y = ln(x − 2)
2
133. y = x + x + 6
112. y = tan(x)
x−1
113. y = 4 − x 2
2
134. y = x − 1
114. y = x 3/2 − 3x 5/2
2
115. y = 2 x − 1
x + 2x − 3
For the following exercises, find the local and/or absolute
maxima for the functions over the specified domain.
f (x) = x 2 + 3 over [−1, 4]
119. y = x 2 + 2x over [1, 4]
2
120. y = ⎛⎝x − x 2⎞⎠ over [−1, 1]
⎛
⎝x
For the following functions, use a calculator to graph the
function and to estimate the absolute and local maxima and
minima. Then, solve for them explicitly.
136. [T] y = x + sin(x)
117. y = x + 1x
121. y =
x−1
135. [T] y = 3x 1 − x 2
116. y = sin 2(x)
118.
132. y = x 3 (1 − x) 6
1
over [0, 1]
− x 2⎞⎠
122. y = 9 − x over [1, 9]
137. [T] y = 12x 5 + 45x 4 + 20x 3 − 90x 2 − 120x + 3
3
2
138. [T] y = x + 6x − x − 30
x−2
2
139. [T] y = 4 − x
4 + x2
140. A company that produces cell phones has a cost
function of C = x 2 − 1200x + 36,400, where C is cost
in dollars and x is number of cell phones produced (in
thousands). How many units of cell phone (in thousands)
minimizes this cost function?
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Chapter 4 | Applications of Derivatives
141. A ball is thrown into the air and its position is given
by h(t) = −4.9t 2 + 60t + 5 m. Find the height at which
the ball stops ascending. How long after it is thrown does
this happen?
For the following exercises, consider the production of
gold during the California gold rush (1848–1888). The
production of gold can be modeled by G(t) =
(25t)
,
+ 16⎞⎠
⎛ 2
⎝t
where t is the number of years since the rush began
(0 ≤ t ≤ 40) and G is ounces of gold produced (in
millions). A summary of the data is shown in the following
figure.
142. Find when the maximum (local and global) gold
production occurred, and the amount of gold produced
during that maximum.
143. Find when the minimum (local and global) gold
production occurred. What was the amount of gold
produced during this minimum?
Find the critical points, maxima, and minima for the
following piecewise functions.
⎧x 2 − 4x 0 ≤ x ≤ 1
144. y = ⎨
⎩ x2 − 4 1 < x ≤ 2
⎧
145. y = ⎨
x2 + 1
x≤1
⎩x 2 − 4x + 5 x > 1
For the following exercises, find the critical points of the
following generic functions. Are they maxima, minima, or
neither? State the necessary conditions.
146. y = ax 2 + bx + c, given that a > 0
147. y = (x − 1) a, given that a > 1
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Chapter 4 | Applications of Derivatives
379
4.4 | The Mean Value Theorem
Learning Objectives
4.4.1 Explain the meaning of Rolle’s theorem.
4.4.2 Describe the significance of the Mean Value Theorem.
4.4.3 State three important consequences of the Mean Value Theorem.
The Mean Value Theorem is one of the most important theorems in calculus. We look at some of its implications at the
end of this section. First, let’s start with a special case of the Mean Value Theorem, called Rolle’s theorem.
Rolle’s Theorem
Informally, Rolle’s theorem states that if the outputs of a differentiable function f are equal at the endpoints of an interval,
then there must be an interior point c where f ′(c) = 0. Figure 4.21 illustrates this theorem.
Figure 4.21 If a differentiable function f satisfies f (a) = f (b), then its derivative must be zero at some point(s)
between a and b.
Theorem 4.4: Rolle’s Theorem
Let f be a continuous function over the closed interval [a, b] and differentiable over the open interval (a, b) such
that f (a) = f (b). There then exists at least one c ∈ (a, b) such that f ′(c) = 0.
Proof
Let k = f (a) = f (b). We consider three cases:
1.
f (x) = k for all x ∈ (a, b).
2. There exists x ∈ (a, b) such that f (x) > k.
3. There exists x ∈ (a, b) such that f (x) < k.
Case 1: If f (x) = 0 for all x ∈ (a, b), then f ′(x) = 0 for all x ∈ (a, b).
Case 2: Since f is a continuous function over the closed, bounded interval [a, b], by the extreme value theorem, it has
an absolute maximum. Also, since there is a point x ∈ (a, b) such that f (x) > k, the absolute maximum is greater than
k. Therefore, the absolute maximum does not occur at either endpoint. As a result, the absolute maximum must occur at an
interior point c ∈ (a, b). Because f has a maximum at an interior point c, and f is differentiable at c, by Fermat’s
theorem, f ′(c) = 0.
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Chapter 4 | Applications of Derivatives
Case 3: The case when there exists a point x ∈ (a, b) such that f (x) < k is analogous to case 2, with maximum replaced
by minimum.
□
An important point about Rolle’s theorem is that the differentiability of the function f is critical. If f is not differentiable,
even at a single point, the result may not hold. For example, the function f (x) = |x| − 1 is continuous over [−1, 1] and
f (−1) = 0 = f (1), but f ′(c) ≠ 0 for any c ∈ (−1, 1) as shown in the following figure.
Figure 4.22 Since f (x) = |x| − 1 is not differentiable at
x = 0, the conditions of Rolle’s theorem are not satisfied. In
fact, the conclusion does not hold here; there is no c ∈ (−1, 1)
such that f ′(c) = 0.
Let’s now consider functions that satisfy the conditions of Rolle’s theorem and calculate explicitly the points c where
f ′(c) = 0.
Example 4.14
Using Rolle’s Theorem
For each of the following functions, verify that the function satisfies the criteria stated in Rolle’s theorem and find
all values c in the given interval where f ′(c) = 0.
a.
f (x) = x 2 + 2x over [−2, 0]
b.
f (x) = x 3 − 4x over [−2, 2]
Solution
a. Since f is a polynomial, it is continuous and differentiable everywhere. In addition, f (−2) = 0 = f (0).
Therefore, f satisfies the criteria of Rolle’s theorem. We conclude that there exists at least one value
c ∈ (−2, 0)
such
that
f ′(c) = 0.
Since
f ′(x) = 2x + 2 = 2(x + 1),
f ′(c) = 2(c + 1) = 0 implies c = −1 as shown in the following graph.
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we
see
that
Chapter 4 | Applications of Derivatives
381
Figure 4.23 This function is continuous and differentiable
over [−2, 0], f ′(c) = 0 when c = −1.
b. As in part a. f is a polynomial and therefore is continuous and differentiable everywhere. Also,
f (−2) = 0 = f (2). That said, f satisfies the criteria of Rolle’s theorem. Differentiating, we find that
f ′(x) = 3x 2 − 4. Therefore, f ′(c) = 0 when x = ± 2 . Both points are in the interval [−2, 2], and,
3
therefore, both points satisfy the conclusion of Rolle’s theorem as shown in the following graph.
Figure 4.24 For this polynomial over [−2, 2], f ′(c) = 0
at x = ±2/ 3.
4.14
Verify that the function f (x) = 2x 2 − 8x + 6 defined over the interval [1, 3] satisfies the conditions of
Rolle’s theorem. Find all points c guaranteed by Rolle’s theorem.
382
Chapter 4 | Applications of Derivatives
The Mean Value Theorem and Its Meaning
Rolle’s theorem is a special case of the Mean Value Theorem. In Rolle’s theorem, we consider differentiable functions
f that are zero at the endpoints. The Mean Value Theorem generalizes Rolle’s theorem by considering functions that are
not necessarily zero at the endpoints. Consequently, we can view the Mean Value Theorem as a slanted version of Rolle’s
theorem (Figure 4.25). The Mean Value Theorem states that if f is continuous over the closed interval [a, b] and
differentiable over the open interval (a, b), then there exists a point c ∈ (a, b) such that the tangent line to the graph of
f at c is parallel to the secant line connecting ⎛⎝a, f (a)⎞⎠ and ⎛⎝b, f (b)⎞⎠.
Figure 4.25 The Mean Value Theorem says that for a function
that meets its conditions, at some point the tangent line has the
same slope as the secant line between the ends. For this
function, there are two values c 1 and c 2 such that the tangent
line to f at c 1 and c 2 has the same slope as the secant line.
Theorem 4.5: Mean Value Theorem
Let f be continuous over the closed interval [a, b] and differentiable over the open interval (a, b). Then, there
exists at least one point c ∈ (a, b) such that
f ′(c) =
f (b) − f (a)
.
b−a
Proof
The proof follows from Rolle’s theorem by introducing an appropriate function that satisfies the criteria of Rolle’s theorem.
Consider the line connecting ⎛⎝a, f (a)⎞⎠ and ⎛⎝b, f (b)⎞⎠. Since the slope of that line is
f (b) − f (a)
b−a
and the line passes through the point ⎛⎝a, f (a)⎞⎠, the equation of that line can be written as
y=
f (b) − f (a)
(x − a) + f (a).
b−a
Let g(x) denote the vertical difference between the point ⎛⎝x, f (x)⎞⎠ and the point (x, y) on that line. Therefore,
⎡ f (b) − f (a)
⎤
(x − a) + f (a)⎦.
g(x) = f (x) − ⎣
b−a
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Chapter 4 | Applications of Derivatives
383
Figure 4.26 The value g(x) is the vertical difference
between the point (x, f (x)) and the point (x, y) on the secant
line connecting (a, f (a)) and (b, f (b)).
Since the graph of f intersects the secant line when x = a and x = b, we see that g(a) = 0 = g(b). Since f is a
differentiable function over (a, b),
over [a, b],
g is also a differentiable function over (a, b). Furthermore, since f is continuous
g is also continuous over [a, b]. Therefore, g satisfies the criteria of Rolle’s theorem. Consequently, there
exists a point c ∈ (a, b) such that g′(c) = 0. Since
g′(x) = f ′(x) −
f (b) − f (a)
,
b−a
g′(c) = f ′(c) −
f (b) − f (a)
.
b−a
we see that
Since g′(c) = 0, we conclude that
f ′(c) =
f (b) − f (a)
.
b−a
□
In the next example, we show how the Mean Value Theorem can be applied to the function f (x) = x over the interval
[0, 9]. The method is the same for other functions, although sometimes with more interesting consequences.
Example 4.15
Verifying that the Mean Value Theorem Applies
For f (x) = x over the interval [0, 9], show that f satisfies the hypothesis of the Mean Value Theorem, and
therefore there exists at least one value c ∈ (0, 9) such that f ′(c) is equal to the slope of the line connecting
⎛
⎝
0, f (0)⎞⎠ and ⎛⎝9, f (9)⎞⎠. Find these values c guaranteed by the Mean Value Theorem.
Solution
We know that f (x) = x is continuous over [0, 9] and differentiable over (0, 9). Therefore, f satisfies the
hypotheses of the Mean Value Theorem, and there must exist at least one value c ∈ (0, 9) such that f ′ (c) is
equal to the slope of the line connecting ⎛⎝0, f (0)⎞⎠ and ⎛⎝9, f (9)⎞⎠ (Figure 4.27). To determine which value(s)
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Chapter 4 | Applications of Derivatives
of c are guaranteed, first calculate the derivative of f . The derivative f ′ (x) =
1 . The slope of the line
(2 x)
connecting (0, f (0)) and (9, f (9)) is given by
f (9) − f (0)
= 9 − 0 = 3 = 1.
9−0
9 3
9−0
We want to find c such that f ′(c) = 1 . That is, we want to find c such that
3
1 = 1.
2 c 3
Solving this equation for c, we obtain c = 9 . At this point, the slope of the tangent line equals the slope of the
4
line joining the endpoints.
Figure 4.27 The slope of the tangent line at c = 9/4 is the same as the slope of the line segment
connecting (0, 0) and (9, 3).
One application that helps illustrate the Mean Value Theorem involves velocity. For example, suppose we drive a car for
1 h down a straight road with an average velocity of 45 mph. Let s(t) and v(t) denote the position and velocity of the
car, respectively, for 0 ≤ t ≤ 1 h. Assuming that the position function s(t) is differentiable, we can apply the Mean Value
Theorem to conclude that, at some time c ∈ (0, 1), the speed of the car was exactly
v(c) = s′ (c) =
s(1) − s(0)
= 45 mph.
1−0
Example 4.16
Mean Value Theorem and Velocity
If a rock is dropped from a height of 100 ft, its position t seconds after it is dropped until it hits the ground is
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Chapter 4 | Applications of Derivatives
385
given by the function s(t) = −16t 2 + 100.
a. Determine how long it takes before the rock hits the ground.
b. Find the average velocity v avg of the rock for when the rock is released and the rock hits the ground.
c. Find the time t guaranteed by the Mean Value Theorem when the instantaneous velocity of the rock is
v avg .
Solution
a. When the rock hits the ground, its position is s(t) = 0. Solving the equation −16t 2 + 100 = 0 for t,
we find that t = ± 5 sec. Since we are only considering t ≥ 0, the ball will hit the ground 5 sec after
2
2
it is dropped.
b. The average velocity is given by
v avg =
s(5/2) − s(0) 1 − 100
=
= −40 ft/sec.
5/2 − 0
5/2
c. The instantaneous velocity is given by the derivative of the position function. Therefore, we need to find
a time t such that v(t) = s′(t) = v avg = −40 ft/sec. Since s(t) is continuous over the interval [0, 5/2]
and differentiable over the interval (0, 5/2), by the Mean Value Theorem, there is guaranteed to be a
point c ∈ (0, 5/2) such that
s′ (c) =
s(5/2) − s(0)
= −40.
5/2 − 0
Taking the derivative of the position function s(t), we find that s′ (t) = −32t. Therefore, the equation
reduces to s′ (c) = −32c = −40. Solving this equation for c, we have c = 5 . Therefore, 5 sec after
4
4
the rock is dropped, the instantaneous velocity equals the average velocity of the rock during its free fall:
−40 ft/sec.
Figure 4.28 At time t = 5/4 sec, the velocity of the rock is
equal to its average velocity from the time it is dropped until it
hits the ground.
4.15
Suppose a ball is dropped from a height of 200 ft. Its position at time t is s(t) = −16t 2 + 200. Find the
time t when the instantaneous velocity of the ball equals its average velocity.
386
Chapter 4 | Applications of Derivatives
Corollaries of the Mean Value Theorem
Let’s now look at three corollaries of the Mean Value Theorem. These results have important consequences, which we use
in upcoming sections.
At this point, we know the derivative of any constant function is zero. The Mean Value Theorem allows us to conclude
that the converse is also true. In particular, if f ′ (x) = 0 for all x in some interval I, then f (x) is constant over that
interval. This result may seem intuitively obvious, but it has important implications that are not obvious, and we discuss
them shortly.
Theorem 4.6: Corollary 1: Functions with a Derivative of Zero
Let f be differentiable over an interval I. If f ′ (x) = 0 for all x ∈ I, then f (x) = constant for all x ∈ I.
Proof
Since f is differentiable over I,
f must be continuous over I. Suppose f (x) is not constant for all x in I. Then there
exist a, b ∈ I, where a ≠ b and f (a) ≠ f (b). Choose the notation so that a < b. Therefore,
f (b) − f (a)
≠ 0.
b−a
Since f is a differentiable function, by the Mean Value Theorem, there exists c ∈ (a, b) such that
f ′ (c) =
f (b) − f (a)
.
b−a
Therefore, there exists c ∈ I such that f ′ (c) ≠ 0, which contradicts the assumption that f ′ (x) = 0 for all x ∈ I.
□
From Corollary 1: Functions with a Derivative of Zero, it follows that if two functions have the same derivative,
they differ by, at most, a constant.
Theorem 4.7: Corollary 2: Constant Difference Theorem
If f and g are differentiable over an interval I and f ′ (x) = g′ (x) for all x ∈ I, then f (x) = g(x) + C for some
constant C.
Proof
Let h(x) = f (x) − g(x). Then, h′ (x) = f ′ (x) − g′ (x) = 0 for all x ∈ I. By Corollary 1, there is a constant C such that
h(x) = C for all x ∈ I. Therefore, f (x) = g(x) + C for all x ∈ I.
□
The third corollary of the Mean Value Theorem discusses when a function is increasing and when it is decreasing. Recall
that a function f is increasing over I if f (x 1) < f (x 2) whenever x 1 < x 2, whereas f is decreasing over I if
f (x) 1 > f (x 2) whenever x 1 < x 2. Using the Mean Value Theorem, we can show that if the derivative of a function is
positive, then the function is increasing; if the derivative is negative, then the function is decreasing (Figure 4.29). We
make use of this fact in the next section, where we show how to use the derivative of a function to locate local maximum
and minimum values of the function, and how to determine the shape of the graph.
This fact is important because it means that for a given function f , if there exists a function F such that F′ (x) = f (x);
then, the only other functions that have a derivative equal to f are F(x) + C for some constant C. We discuss this result
in more detail later in the chapter.
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Chapter 4 | Applications of Derivatives
387
Figure 4.29 If a function has a positive derivative over some interval I, then the function
increases over that interval I; if the derivative is negative over some interval I, then the
function decreases over that interval I.
Theorem 4.8: Corollary 3: Increasing and Decreasing Functions
Let f be continuous over the closed interval [a, b] and differentiable over the open interval (a, b).
i. If f ′ (x) > 0 for all x ∈ (a, b), then f is an increasing function over [a, b].
ii. If f ′(x) < 0 for all x ∈ (a, b), then f is a decreasing function over [a, b].
Proof
We will prove i.; the proof of ii. is similar. Suppose f is not an increasing function on I. Then there exist a and b in I
such that a < b, but f (a) ≥ f (b). Since f is a differentiable function over I, by the Mean Value Theorem there exists
c ∈ (a, b) such that
f ′ (c) =
f (b) − f (a)
.
b−a
Since f (a) ≥ f (b), we know that f (b) − f (a) ≤ 0. Also, a < b tells us that b − a > 0. We conclude that
f ′ (c) =
f (b) − f (a)
≤ 0.
b−a
However, f ′ (x) > 0 for all x ∈ I. This is a contradiction, and therefore f must be an increasing function over I.
□
388
Chapter 4 | Applications of Derivatives
4.4 EXERCISES
148. Why do you need continuity to apply the Mean Value
Theorem? Construct a counterexample.
163.
f (x) = cos(2πx)
149. Why do you need differentiability to apply the Mean
Value Theorem? Find a counterexample.
164.
f (x) = 1 + x + x 2
165.
f (x) = (x − 1) 10
166.
f (x) = (x − 1) 9
150. When are Rolle’s theorem and the Mean Value
Theorem equivalent?
151. If you have a function with a discontinuity, is it still
possible to have f ′ (c)(b − a) = f (b) − f (a) ? Draw such
an example or prove why not.
For the following exercises, determine over what intervals
(if any) the Mean Value Theorem applies. Justify your
answer.
152. y = sin(πx)
153. y = 13
x
154. y = 4 − x 2
| |
167.
f (x) = x − 1
2
168.
f (x) = 12
x
169.
f (x) = |x|
170.
f (x) = ⌊x⌋ (Hint: This is called the floor function
or equal to x.)
156. y = ln(3x − 5)
For the following exercises, graph the functions on a
calculator and draw the secant line that connects the
endpoints. Estimate the number of points c such that
f ′(c)(b − a) = f (b) − f (a).
For the following exercises, determine whether the Mean
Value Theorem applies for the functions over the given
interval [a, b]. Justify your answer.
171. y = e x over [0, 1]
⎡
⎤
172. y = ln(2x + 3) over ⎣− 3 , 0⎦
2
157. [T] y = 3x 3 + 2x + 1 over [−1, 1]
⎡
⎤
⎛ ⎞
158. [T] y = tan⎝π x⎠ over ⎣− 3 , 3 ⎦
4
2 2
173.
f (x) = tan(2πx) over [0, 2]
174. y = 9 − x 2 over [−3, 3]
2
159. [T] y = x cos(πx) over [−2, 2]
y = x 6 − 3 x 5 − 9 x 4 + 15 x 3 + 3 x 2 + 3 x + 1
4
8
32
32
16
16
[−1, 1]
Theorem does not apply over the interval [−1, 1].
and it is defined so that f (x) is the largest integer less than
155. y = x 2 − 4
160.
For the following exercises, show there is no c such that
f (1) − f (−1) = f ′ (c)(2). Explain why the Mean Value
[T]
over
For the following exercises, use the Mean Value Theorem
and
find
all
points
0<c<2
such
that
f (2) − f (0) = f ′ (c)(2 − 0).
175. y =
1
| x + 1|
over [0, 3]
176. y = x 3 + 2x + 1 over [0, 6]
2
+ 2 over [−1, 1]
177. y = x + 3x
x
178. y =
x
over [0, 1]
sin(πx) + 1
161.
f (x) = x 3
179. y = ln(x + 1) over [0, e − 1]
162.
f (x) = sin(πx)
180. y = x sin(πx) over [0, 2]
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Chapter 4 | Applications of Derivatives
181. y = 5 + |x| over [−1, 1]
For the following exercises, consider the roots of the
equation.
182.
Show that the equation y = x 3 + 3x 2 + 16 has
exactly one real root. What is it?
183. Find the conditions for exactly one root (double root)
for the equation y = x 2 + bx + c
184. Find the conditions for y = e x − b to have one root.
Is it possible to have more than one root?
For the following exercises, use a calculator to graph the
function over the interval [a, b] and graph the secant line
from a to b. Use the calculator to estimate all values of c
as guaranteed by the Mean Value Theorem. Then, find the
exact value of c, if possible, or write the final equation
and use a calculator to estimate to four digits.
⎡
⎤
185. [T] y = tan(πx) over ⎣− 1 , 1 ⎦
4 4
1
over [0, 3]
x+1
186. [T] y =
|
|
187. [T] y = x 2 + 2x − 4 over [−4, 0]
⎡
⎤
188. [T] y = x + 1x over ⎣ 1 , 4⎦
2
189. [T] y = x + 1 + 12 over [3, 8]
x
190. At 10:17 a.m., you pass a police car at 55 mph that
is stopped on the freeway. You pass a second police car at
55 mph at 10:53 a.m., which is located 39 mi from the first
police car. If the speed limit is 60 mph, can the police cite
you for speeding?
191. Two cars drive from one spotlight to the next, leaving
at the same time and arriving at the same time. Is there
ever a time when they are going the same speed? Prove or
disprove.
192. Show that y = sec 2 x and y = tan 2 x have the same
derivative. What can you say about y = sec 2 x − tan 2 x ?
193. Show that y = csc 2 x and y = cot 2 x have the same
derivative. What can you say about y = csc 2 x − cot 2 x ?
389
390
Chapter 4 | Applications of Derivatives
4.5 | Derivatives and the Shape of a Graph
Learning Objectives
4.5.1 Explain how the sign of the first derivative affects the shape of a function’s graph.
4.5.2 State the first derivative test for critical points.
4.5.3 Use concavity and inflection points to explain how the sign of the second derivative affects
the shape of a function’s graph.
4.5.4 Explain the concavity test for a function over an open interval.
4.5.5 Explain the relationship between a function and its first and second derivatives.
4.5.6 State the second derivative test for local extrema.
Earlier in this chapter we stated that if a function f has a local extremum at a point c, then c must be a critical point
of f . However, a function is not guaranteed to have a local extremum at a critical point. For example, f (x) = x 3 has a
critical point at x = 0 since f ′(x) = 3x 2 is zero at x = 0, but f does not have a local extremum at x = 0. Using the
results from the previous section, we are now able to determine whether a critical point of a function actually corresponds
to a local extreme value. In this section, we also see how the second derivative provides information about the shape of a
graph by describing whether the graph of a function curves upward or curves downward.
The First Derivative Test
Corollary 3 of the Mean Value Theorem showed that if the derivative of a function is positive over an interval I then the
function is increasing over I. On the other hand, if the derivative of the function is negative over an interval I, then the
function is decreasing over I as shown in the following figure.
Figure 4.30 Both functions are increasing over the interval
(a, b). At each point x, the derivative f ′(x) > 0. Both
functions are decreasing over the interval (a, b). At each point
x, the derivative f ′(x) < 0.
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Chapter 4 | Applications of Derivatives
391
A continuous function f has a local maximum at point c if and only if f switches from increasing to decreasing at
point c. Similarly, f has a local minimum at c if and only if f switches from decreasing to increasing at c. If f is a
continuous function over an interval I containing c and differentiable over I, except possibly at c, the only way f
can switch from increasing to decreasing (or vice versa) at point c is if f ′ changes sign as x increases through c. If
f is differentiable at c, the only way that f ′. can change sign as x increases through c is if f ′ (c) = 0. Therefore,
for a function f that is continuous over an interval I containing c and differentiable over I, except possibly at c, the
only way f can switch from increasing to decreasing (or vice versa) is if f ′(c) = 0 or f ′ (c) is undefined. Consequently,
to locate local extrema for a function f , we look for points c in the domain of f such that f ′(c) = 0 or f ′ (c) is
undefined. Recall that such points are called critical points of f .
Note that f need not have a local extrema at a critical point. The critical points are candidates for local extrema only. In
Figure 4.31, we show that if a continuous function f has a local extremum, it must occur at a critical point, but a function
may not have a local extremum at a critical point. We show that if f has a local extremum at a critical point, then the sign
of f ′ switches as x increases through that point.
Figure 4.31 The function f has four critical points: a, b, c, and d. The function f has local maxima at a
and d, and a local minimum at b. The function f does not have a local extremum at c. The sign of f ′
changes at all local extrema.
Using Figure 4.31, we summarize the main results regarding local extrema.
• If a continuous function f has a local extremum, it must occur at a critical point c.
• The function has a local extremum at the critical point c if and only if the derivative f ′ switches sign as x
increases through c.
• Therefore, to test whether a function has a local extremum at a critical point c, we must determine the sign of
f ′ (x) to the left and right of c.
This result is known as the first derivative test.
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Chapter 4 | Applications of Derivatives
Theorem 4.9: First Derivative Test
Suppose that f is a continuous function over an interval I containing a critical point c. If f is differentiable over
I, except possibly at point c, then f (c) satisfies one of the following descriptions:
i. If f ′ changes sign from positive when x < c to negative when x > c, then f (c) is a local maximum of f .
ii. If f ′ changes sign from negative when x < c to positive when x > c, then f (c) is a local minimum of f .
iii. If f ′ has the same sign for x < c and x > c, then f (c) is neither a local maximum nor a local minimum of
f.
We can summarize the first derivative test as a strategy for locating local extrema.
Problem-Solving Strategy: Using the First Derivative Test
Consider a function f that is continuous over an interval I.
1. Find all critical points of f and divide the interval I into smaller intervals using the critical points as
endpoints.
2. Analyze the sign of f ′ in each of the subintervals. If f ′ is continuous over a given subinterval (which is
typically the case), then the sign of f ′ in that subinterval does not change and, therefore, can be determined
by choosing an arbitrary test point x in that subinterval and by evaluating the sign of f ′ at that test point. Use
the sign analysis to determine whether f is increasing or decreasing over that interval.
3. Use First Derivative Test and the results of step 2 to determine whether f has a local maximum, a local
minimum, or neither at each of the critical points.
Now let’s look at how to use this strategy to locate all local extrema for particular functions.
Example 4.17
Using the First Derivative Test to Find Local Extrema
Use the first derivative test to find the location of all local extrema for f (x) = x 3 − 3x 2 − 9x − 1. Use a
graphing utility to confirm your results.
Solution
Step 1. The derivative is f ′ (x) = 3x 2 − 6x − 9. To find the critical points, we need to find where f ′ (x) = 0.
Factoring the polynomial, we conclude that the critical points must satisfy
3(x 2 − 2x − 3) = 3(x − 3)(x + 1) = 0.
Therefore, the critical points are x = 3, −1. Now divide the interval (−∞, ∞) into the smaller intervals
(−∞, −1), (−1, 3) and (3, ∞).
Step 2. Since f ′ is a continuous function, to determine the sign of f ′ (x) over each subinterval, it suffices to
choose a point over each of the intervals (−∞, −1), (−1, 3) and (3, ∞) and determine the sign of f ′ at each
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Chapter 4 | Applications of Derivatives
393
of these points. For example, let’s choose x = −2, x = 0, and x = 4 as test points.
Interval
Test Point
Sign of f′ (x) = 3(x − 3)(x + 1) at Test Point
(−∞, −1)
x = −2
(+)(−)(−) = +
f is increasing.
(−1, 3)
x=0
(+)(−)(+) = −
f is decreasing.
(3, ∞)
x=4
(+)(+)(+) = +
f is increasing.
Conclusion
Step 3. Since f ′ switches sign from positive to negative as x increases through 1, f has a local maximum at
x = −1. Since f ′ switches sign from negative to positive as x increases through 3, f has a local minimum at
x = 3. These analytical results agree with the following graph.
Figure 4.32 The function f has a maximum at x = −1 and
a minimum at x = 3
394
Chapter 4 | Applications of Derivatives
4.16
Use the first derivative test to locate all local extrema for f (x) = −x 3 + 3 x 2 + 18x.
2
Example 4.18
Using the First Derivative Test
Use the first derivative test to find the location of all local extrema for f (x) = 5x 1/3 − x 5/3. Use a graphing
utility to confirm your results.
Solution
Step 1. The derivative is
5⎛⎝1 − x 4/3⎞⎠
2/3
4/3
f ′ (x) = 5 x −2/3 − 5 x 2/3 = 52/3 − 5x = 5 − 5x
=
.
3
3
3
3x 2/3
3x
3x 2/3
The derivative f ′ (x) = 0 when 1 − x 4/3 = 0. Therefore, f ′ (x) = 0 at x = ±1. The derivative f ′ (x) is
undefined at x = 0. Therefore, we have three critical points: x = 0,
x = 1, and x = −1. Consequently,
divide the interval (−∞, ∞) into the smaller intervals (−∞, −1), (−1, 0), (0, 1), and (1, ∞).
Step 2: Since f ′ is continuous over each subinterval, it suffices to choose a test point x in each of the
intervals from step
1
and determine the sign of
f′
at each of these points. The points
x = −2, x = − 1 , x = 1 , and x = 2 are test points for these intervals.
2
2
Interval
Test Point
Sign of f′ (x) =
5⎛⎝1 − x4/3⎞⎠
3x2/3
at Test Point
Conclusion
(−∞, −1)
x = −2
(+)(−)
+ =−
f is decreasing.
(−1, 0)
x= −1
2
(+)(+)
+ = +
f is increasing.
(0, 1)
x=1
2
(+)(+)
+ = +
f is increasing.
(1, ∞)
x=2
(+)(−)
+ =−
f is decreasing.
Step 3: Since f is decreasing over the interval (−∞, −1) and increasing over the interval (−1, 0),
local minimum at x = −1. Since f is increasing over the interval (−1, 0) and the interval (0, 1),
f has a
f does not
have a local extremum at x = 0. Since f is increasing over the interval (0, 1) and decreasing over the interval
(1, ∞), f has a local maximum at x = 1. The analytical results agree with the following graph.
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Chapter 4 | Applications of Derivatives
395
Figure 4.33 The function f has a local minimum at x = −1
and a local maximum at x = 1.
4.17
3
Use the first derivative test to find all local extrema for f (x) = x − 1.
Concavity and Points of Inflection
We now know how to determine where a function is increasing or decreasing. However, there is another issue to consider
regarding the shape of the graph of a function. If the graph curves, does it curve upward or curve downward? This notion is
called the concavity of the function.
Figure 4.34(a) shows a function f with a graph that curves upward. As x increases, the slope of the tangent line
increases. Thus, since the derivative increases as x increases, f ′ is an increasing function. We say this function f is
concave up. Figure 4.34(b) shows a function f that curves downward. As x increases, the slope of the tangent line
decreases. Since the derivative decreases as x increases, f ′ is a decreasing function. We say this function f is concave
down.
Definition
Let f be a function that is differentiable over an open interval I. If f ′ is increasing over I, we say f is concave
up over I. If f ′ is decreasing over I, we say f is concave down over I.
396
Chapter 4 | Applications of Derivatives
Figure 4.34 (a), (c) Since f ′ is increasing over the interval (a, b), we say f
is concave up over (a, b). (b), (d) Since f ′ is decreasing over the interval
(a, b), we say f is concave down over (a, b).
In general, without having the graph of a function f , how can we determine its concavity? By definition, a function f is
concave up if f ′ is increasing. From Corollary 3, we know that if f ′ is a differentiable function, then f ′ is increasing
if its derivative f ″(x) > 0. Therefore, a function f that is twice differentiable is concave up when f ″(x) > 0. Similarly,
a function f is concave down if f ′ is decreasing. We know that a differentiable function f ′ is decreasing if its derivative
f ″(x) < 0. Therefore, a twice-differentiable function f is concave down when f ″(x) < 0. Applying this logic is known
as the concavity test.
Theorem 4.10: Test for Concavity
Let f be a function that is twice differentiable over an interval I.
i. If f ″(x) > 0 for all x ∈ I, then f is concave up over I.
ii. If f ″(x) < 0 for all x ∈ I, then f is concave down over I.
We conclude that we can determine the concavity of a function f by looking at the second derivative of f . In addition, we
observe that a function f can switch concavity (Figure 4.35). However, a continuous function can switch concavity only
at a point x if f ″(x) = 0 or f ″(x) is undefined. Consequently, to determine the intervals where a function f is concave
up and concave down, we look for those values of x where f ″(x) = 0 or f ″(x) is undefined. When we have determined
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Chapter 4 | Applications of Derivatives
397
these points, we divide the domain of f into smaller intervals and determine the sign of f ″ over each of these smaller
intervals. If f ″ changes sign as we pass through a point x, then f changes concavity. It is important to remember that a
function f may not change concavity at a point x even if f ″(x) = 0 or f ″(x) is undefined. If, however, f does change
concavity at a point a and f is continuous at a, we say the point ⎛⎝a, f (a)⎞⎠ is an inflection point of f .
Definition
If f is continuous at a and f changes concavity at a, the point ⎛⎝a, f (a)⎞⎠ is an inflection point of f .
Figure 4.35 Since f ″(x) > 0 for x < a, the function f is concave up over the interval
(−∞, a). Since f ″(x) < 0 for x > a, the function f is concave down over the interval
(a, ∞). The point ⎛⎝a, f (a)⎞⎠ is an inflection point of f .
Example 4.19
Testing for Concavity
For the function f (x) = x 3 − 6x 2 + 9x + 30, determine all intervals where f is concave up and all intervals
where f is concave down. List all inflection points for f . Use a graphing utility to confirm your results.
Solution
To determine concavity, we need to find the second derivative
f ″(x).
The first derivative is
f ′(x) = 3x 2 − 12x + 9, so the second derivative is f ″(x) = 6x − 12. If the function changes concavity, it
occurs either when f ″(x) = 0 or f ″(x) is undefined. Since f ″ is defined for all real numbers x, we need only
find where f ″(x) = 0. Solving the equation 6x − 12 = 0, we see that x = 2 is the only place where f could
change concavity. We now test points over the intervals (−∞, 2) and (2, ∞) to determine the concavity of f .
The points x = 0 and x = 3 are test points for these intervals.
398
Chapter 4 | Applications of Derivatives
Sign of f″(x) = 6x − 12 at Test Point
Interval
Test Point
(−∞, 2)
x=0
−
f is concave down
(2, ∞)
x=3
+
f is concave up.
Conclusion
We conclude that f is concave down over the interval (−∞, 2) and concave up over the interval (2, ∞). Since
f changes concavity at x = 2, the point ⎛⎝2, f (2)⎞⎠ = (2, 32) is an inflection point. Figure 4.36 confirms the
analytical results.
Figure 4.36 The given function has a point of inflection at
(2, 32) where the graph changes concavity.
For f (x) = −x 3 + 3 x 2 + 18x, find all intervals where f is concave up and all intervals where f is
2
concave down.
4.18
We now summarize, in Table 4.6, the information that the first and second derivatives of a function f provide about the
graph of f , and illustrate this information in Figure 4.37.
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Chapter 4 | Applications of Derivatives
399
Sign of f′
Sign of f″
Is f increasing or decreasing?
Concavity
Positive
Positive
Increasing
Concave up
Positive
Negative
Increasing
Concave down
Negative
Positive
Decreasing
Concave up
Negative
Negative
Decreasing
Concave down
Table 4.6 What Derivatives Tell Us about Graphs
Figure 4.37 Consider a twice-differentiable function f over an open interval I. If f ′(x) > 0 for all x ∈ I, the
function is increasing over I. If f ′(x) < 0 for all x ∈ I, the function is decreasing over I. If f ″(x) > 0 for all
x ∈ I, the function is concave up. If f ″(x) < 0 for all x ∈ I, the function is concave down on I.
The Second Derivative Test
The first derivative test provides an analytical tool for finding local extrema, but the second derivative can also be used to
locate extreme values. Using the second derivative can sometimes be a simpler method than using the first derivative.
We know that if a continuous function has a local extrema, it must occur at a critical point. However, a function need not
have a local extrema at a critical point. Here we examine how the second derivative test can be used to determine whether
a function has a local extremum at a critical point. Let f be a twice-differentiable function such that f ′ (a) = 0 and f ″
is continuous over an open interval I containing a. Suppose f ″(a) < 0. Since f ″ is continuous over I,
all x ∈ I (Figure 4.38). Then, by Corollary 3,
f ″(x) < 0 for
f ′ is a decreasing function over I. Since f ′ (a) = 0, we conclude that
for all x ∈ I, f ′ (x) > 0 if x < a and f ′ (x) < 0 if x > a. Therefore, by the first derivative test, f has a local maximum
at x = a. On the other hand, suppose there exists a point b such that f ′ (b) = 0 but f ″(b) > 0. Since f ″ is continuous
over an open interval I containing b, then f ″(x) > 0 for all x ∈ I (Figure 4.38). Then, by Corollary 3, f ′ is an
increasing function over I. Since f ′ (b) = 0, we conclude that for all x ∈ I,
x > b. Therefore, by the first derivative test, f has a local minimum at x = b.
f ′ (x) < 0 if x < b and f ′ (x) > 0 if
400
Chapter 4 | Applications of Derivatives
Figure 4.38 Consider a twice-differentiable function f such
that f ″ is continuous. Since f ′(a) = 0 and f ″(a) < 0,
there is an interval I containing a such that for all x in I, f
is increasing if x < a and f is decreasing if x > a. As a
result, f has a local maximum at x = a. Since f ′(b) = 0
and f ″(b) > 0, there is an interval I containing b such that
for all x in I, f is decreasing if x < b and f is increasing
if x > b. As a result, f has a local minimum at x = b.
Theorem 4.11: Second Derivative Test
Suppose f ′ (c) = 0, f ″ is continuous over an interval containing c.
i. If f ″(c) > 0, then f has a local minimum at c.
ii. If f ″(c) < 0, then f has a local maximum at c.
iii. If f ″(c) = 0, then the test is inconclusive.
Note that for case iii. when f ″(c) = 0, then f may have a local maximum, local minimum, or neither at c. For
example, the functions f (x) = x 3,
f (x) = x 4, and f (x) = −x 4 all have critical points at x = 0. In each case, the
second derivative is zero at x = 0. However, the function f (x) = x 4 has a local minimum at x = 0 whereas the function
f (x) = −x 4 has a local maximum at x, and the function f (x) = x 3 does not have a local extremum at x = 0.
Let’s now look at how to use the second derivative test to determine whether f has a local maximum or local minimum at
a critical point c where f ′ (c) = 0.
Example 4.20
Using the Second Derivative Test
Use the second derivative to find the location of all local extrema for f (x) = x 5 − 5x 3.
Solution
To apply the second derivative test, we first need to find critical points c where f ′ (c) = 0. The derivative is
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Chapter 4 | Applications of Derivatives
401
f ′ (x) = 5x 4 − 15x 2. Therefore, f ′ (x) = 5x 4 − 15x 2 = 5x 2 ⎛⎝x 2 − 3⎞⎠ = 0 when x = 0, ± 3.
To determine whether f has a local extrema at any of these points, we need to evaluate the sign of f ″ at these
points. The second derivative is
f ″(x) = 20x 3 − 30x = 10x⎛⎝2x 2 − 3⎞⎠.
In the following table, we evaluate the second derivative at each of the critical points and use the second
derivative test to determine whether f has a local maximum or local minimum at any of these points.
x
f″(x)
Conclusion
− 3
−30 3
Local maximum
0
0
Second derivative test is inconclusive
30 3
Local minimum
3
By the second derivative test, we conclude that f has a local maximum at x = − 3 and f has a local minimum
at x = 3. The second derivative test is inconclusive at x = 0. To determine whether f has a local extrema at
x = 0, we apply the first derivative test. To evaluate the sign of f ′ (x) = 5x 2 ⎛⎝x 2 − 3⎞⎠ for x ∈ ⎛⎝− 3, 0⎞⎠ and
x ∈ ⎛⎝0, 3⎞⎠, let x = −1 and x = 1 be the two test points. Since f ′ (−1) < 0 and f ′ (1) < 0, we conclude
that f is decreasing on both intervals and, therefore, f does not have a local extrema at x = 0 as shown in the
following graph.
402
Chapter 4 | Applications of Derivatives
Figure 4.39 The function f has a local maximum at x = − 3 and a local minimum at x = 3
⎛ ⎞
Consider the function f (x) = x 3 − ⎝3 ⎠x 2 − 18x. The points c = 3, −2 satisfy f ′ (c) = 0. Use the
2
second derivative test to determine whether f has a local maximum or local minimum at those points.
4.19
We have now developed the tools we need to determine where a function is increasing and decreasing, as well as acquired
an understanding of the basic shape of the graph. In the next section we discuss what happens to a function as x → ±∞.
At that point, we have enough tools to provide accurate graphs of a large variety of functions.
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Chapter 4 | Applications of Derivatives
403
4.5 EXERCISES
194. If c is a critical point of f (x), when is there no
202.
local maximum or minimum at c ? Explain.
195.
For the function y = x 3,
is x = 0 both an
inflection point and a local maximum/minimum?
196. For the function y = x 3, is x = 0 an inflection
point?
197. Is it possible for a point c to be both an inflection
point and a local extrema of a twice differentiable
function?
198. Why do you need continuity for the first derivative
test? Come up with an example.
199. Explain whether a concave-down function has to
cross y = 0 for some value of x.
200. Explain whether a polynomial of degree 2 can have
an inflection point.
203.
For the following exercises, analyze the graphs of f ′,
then list all intervals where f is increasing or decreasing.
201.
204.
404
Chapter 4 | Applications of Derivatives
205.
209.
For the following exercises, analyze the graphs of f ′,
then list all intervals where
a.
f is increasing and decreasing and
210.
b. the minima and maxima are located.
206.
207.
For the following exercises, analyze the graphs of f ′,
then list all inflection points and intervals f that are
concave up and concave down.
211.
208.
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Chapter 4 | Applications of Derivatives
405
212.
215.
213.
For the following exercises, draw a graph that satisfies
the given specifications for the domain x = [−3, 3]. The
function does not have to be continuous or differentiable.
216.
f (x) > 0, f ′ (x) > 0
over
x > 1, −3 < x < 0, f ′ (x) = 0 over 0 < x < 1
217.
f ′ (x) > 0 over x > 2, −3 < x < −1, f ′ (x) < 0
over −1 < x < 2, f ″(x) < 0 for all x
f ″(x) < 0
218.
over
−1 < x < 1, f ″(x) > 0, −3 < x < −1, 1 < x < 3,
local maximum at x = 0, local minima at x = ±2
214.
219. There is a local maximum at x = 2, local minimum
at x = 1, and the graph is neither concave up nor concave
down.
220. There are local maxima at x = ±1, the function is
concave up for all x, and the function remains positive for
all x.
For the following exercises, determine
a. intervals where f is increasing or decreasing and
b. local minima and maxima of f .
221.
f (x) = sin x + sin 3 x over −π < x < π
222.
f (x) = x 2 + cos x
For the following exercises, determine a. intervals where f
is concave up or concave down, and b. the inflection points
of f .
223.
f (x) = x 3 − 4x 2 + x + 2
406
Chapter 4 | Applications of Derivatives
For the following exercises, determine
238.
f (x) = ln x x, x > 0
239.
f (x) = 1 x + 1x , x > 0
4
240.
x
f (x) = ex , x ≠ 0
a. intervals where f is increasing or decreasing,
b. local minima and maxima of f ,
c. intervals where f is concave up and concave
down, and
d. the inflection points of f .
For the following exercises, interpret the sentences in terms
of f , f ′, and f ″.
224.
f (x) = x 2 − 6x
225.
f (x) = x 3 − 6x 2
the population.
226.
f (x) = x 4 − 6x 3
242. A bike accelerates faster, but a car goes faster. Here
f = Bike’s position minus Car’s position.
227.
f (x) = x 11 − 6x 10
243. The airplane lands smoothly. Here f is the plane’s
2
228.
f (x) = x + x − x
229.
f (x) = x 2 + x + 1
230.
f (x) = x 3 + x 4
241. The population is growing more slowly. Here f is
altitude.
3
244. Stock prices are at their peak. Here f is the stock
price.
245. The economy is picking up speed. Here f is a
measure of the economy, such as GDP.
a. intervals where f is increasing or decreasing,
For the following exercises, consider a third-degree
polynomial
f (x),
which has the properties
b. local minima and maxima of f ,
statements are true or false. Justify your answer.
For the following exercises, determine
c. intervals where f is concave up and concave
down, and
d. the inflection points of f . Sketch the curve, then
use a calculator to compare your answer. If you
cannot determine the exact answer analytically, use
a calculator.
f ′ (1) = 0, f ′ (3) = 0. Determine whether the following
246.
f (x) = 0 for some 1 ≤ x ≤ 3
247.
f ″(x) = 0 for some 1 ≤ x ≤ 3
248. There is no absolute maximum at x = 3
249. If f (x) has three roots, then it has 1 inflection
231. [T] f (x) = sin(πx) − cos(πx) over x = [−1, 1]
point.
⎤
⎡
232. [T] f (x) = x + sin(2x) over x = ⎣− π , π ⎦
2 2
250. If f (x) has one inflection point, then it has three real
⎛
⎞
233. [T] f (x) = sin x + tan x over ⎝− π , π ⎠
2 2
234. [T] f (x) = (x − 2) 2 (x − 4) 2
235. [T] f (x) =
1 ,x≠1
1−x
x
f (x) = sin
x
[2π, 0) ∪ (0, 2π]
236.
237.
[T]
over
x = [−2π, 2π]
f (x) = sin(x)e x over x = [−π, π]
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roots.
Chapter 4 | Applications of Derivatives
407
4.6 | Limits at Infinity and Asymptotes
Learning Objectives
4.6.1 Calculate the limit of a function as x increases or decreases without bound.
4.6.2 Recognize a horizontal asymptote on the graph of a function.
4.6.3 Estimate the end behavior of a function as x increases or decreases without bound.
4.6.4 Recognize an oblique asymptote on the graph of a function.
4.6.5 Analyze a function and its derivatives to draw its graph.
We have shown how to use the first and second derivatives of a function to describe the shape of a graph. To graph a
function f defined on an unbounded domain, we also need to know the behavior of f as x → ±∞. In this section, we
define limits at infinity and show how these limits affect the graph of a function. At the end of this section, we outline a
strategy for graphing an arbitrary function f .
Limits at Infinity
We begin by examining what it means for a function to have a finite limit at infinity. Then we study the idea of a function
with an infinite limit at infinity. Back in Introduction to Functions and Graphs, we looked at vertical asymptotes; in
this section we deal with horizontal and oblique asymptotes.
Limits at Infinity and Horizontal Asymptotes
f (x) = L means f (x) becomes arbitrarily close to L as long as x is sufficiently close to a. We can
Recall that xlim
→a
extend this idea to limits at infinity. For example, consider the function f (x) = 2 + 1x . As can be seen graphically in
Figure 4.40 and numerically in Table 4.8, as the values of x get larger, the values of f (x) approach 2. We say the limit
as x approaches ∞ of f (x) is 2 and write x lim
f (x) = 2. Similarly, for x < 0, as the values |x| get larger, the values
→∞
of f (x) approaches 2. We say the limit as x approaches −∞ of f (x) is 2 and write xlim
f (x) = 2.
→a
Figure 4.40 The function approaches the asymptote y = 2 as x approaches ±∞.
408
Chapter 4 | Applications of Derivatives
x
10
100
1,000
10,000
2 + 1x
2.1
2.01
2.001
2.0001
x
−10
−100
−1000
−10,000
2 + 1x
1.9
1.99
1.999
1.9999
Table 4.8 Values of a function f as x → ±∞
More generally, for any function f , we say the limit as x → ∞ of f (x) is L if f (x) becomes arbitrarily close to
L as long as x is sufficiently large. In that case, we write xlim
f (x) = L. Similarly, we say the limit as x → −∞ of
→a
f (x) is L if f (x) becomes arbitrarily close to L as long as x < 0 and |x| is sufficiently large. In that case, we write
lim f (x) = L. We now look at the definition of a function having a limit at infinity.
x → −∞
Definition
(Informal) If the values of f (x) become arbitrarily close to L as x becomes sufficiently large, we say the function
f has a limit at infinity and write
lim f (x) = L.
x→∞
If the values of f (x) becomes arbitrarily close to L for x < 0 as |x| becomes sufficiently large, we say that the
function f has a limit at negative infinity and write
lim f (x) = L.
x→∞
If the values f (x) are getting arbitrarily close to some finite value L as x → ∞ or x → −∞, the graph of f approaches
the line y = L. In that case, the line y = L is a horizontal asymptote of f (Figure 4.41). For example, for the function
f (x) = 1x , since x lim
f (x) = 0, the line y = 0 is a horizontal asymptote of f (x) = 1x .
→∞
Definition
If x lim
f (x) = L or x →
lim
f (x) = L, we say the line y = L is a horizontal asymptote of f .
→∞
−∞
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Chapter 4 | Applications of Derivatives
409
Figure 4.41 (a) As x → ∞, the values of f are getting arbitrarily close to L. The line y = L
is a horizontal asymptote of f . (b) As x → −∞, the values of f are getting arbitrarily close to
M. The line y = M is a horizontal asymptote of f .
A function cannot cross a vertical asymptote because the graph must approach infinity (or −∞) from at least one direction
as x approaches the vertical asymptote. However, a function may cross a horizontal asymptote. In fact, a function may
cross a horizontal asymptote an unlimited number of times. For example, the function f (x) =
(cos x)
x + 1 shown in Figure
4.42 intersects the horizontal asymptote y = 1 an infinite number of times as it oscillates around the asymptote with everdecreasing amplitude.
Figure 4.42 The graph of f (x) = (cos x)/x + 1 crosses its
horizontal asymptote y = 1 an infinite number of times.
The algebraic limit laws and squeeze theorem we introduced in Introduction to Limits also apply to limits at infinity. We
illustrate how to use these laws to compute several limits at infinity.
Example 4.21
Computing Limits at Infinity
For each of the following functions f ,
asymptote(s) for f .
a.
f (x) = 5 − 22
x
b.
x
f (x) = sin
x
c.
f (x) = tan −1 (x)
evaluate x lim
f (x) and x →
lim
f (x). Determine the horizontal
→∞
−∞
410
Chapter 4 | Applications of Derivatives
Solution
a. Using
the
algebraic
limit
⎛
⎞
1 ⎞.⎛ lim 1 ⎞ = 5 − 2 · 0 = 5.
5 − 2⎛⎝x lim
lim 5 − 22 = x lim
→∞
→ ∞ x ⎠ ⎝x → ∞ x ⎠
x → ∞⎝
x ⎠
laws,
we
have
Similarly, x lim
f (x) = 5. Therefore, f (x) = 5 − 22 has a horizontal asymptote of y = 5 and f
→∞
x
approaches this horizontal asymptote as x → ±∞ as shown in the following graph.
Figure 4.43 This function approaches a horizontal asymptote
as x → ±∞.
b. Since −1 ≤ sin x ≤ 1 for all x, we have
−1 ≤ sin x ≤ 1
x
x
x
for all x ≠ 0. Also, since
1,
lim −1
x = 0 = x lim
→ ∞x
x→∞
we can apply the squeeze theorem to conclude that
x
lim sin
x = 0.
x→∞
Similarly,
x
lim sin
x = 0.
x → −∞
x
Thus, f (x) = sin
x has a horizontal asymptote of y = 0 and f (x) approaches this horizontal asymptote
as x → ±∞ as shown in the following graph.
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Chapter 4 | Applications of Derivatives
411
Figure 4.44 This function crosses its horizontal asymptote multiple times.
c. To evaluate x lim
tan −1 (x) and x →
lim
tan −1 (x), we first consider the graph of y = tan(x) over the
→∞
−∞
interval (−π/2, π/2) as shown in the following graph.
Figure 4.45 The graph of tan x has vertical asymptotes at
x = ±π
2
Since
lim tan x
x → (π/2)−
= ∞,
it follows that
lim tan −1(x) = π .
2
x→∞
Similarly, since
lim
x → (π/2)+
tan x = −∞,
it follows that
lim tan −1 (x) = − π .
2
x → −∞
As a result, y = π and y = − π are horizontal asymptotes of f (x) = tan −1 (x) as shown in the following
2
graph.
2
412
Chapter 4 | Applications of Derivatives
Figure 4.46 This function has two horizontal asymptotes.
4.20
⎛
⎞
⎛
⎞
Evaluate x →
3 + 4x ⎠ and x lim
3 + 4x ⎠. Determine the horizontal asymptotes of f (x) = 3 + 4x , if
lim
−∞⎝
→ ∞⎝
any.
Infinite Limits at Infinity
Sometimes the values of a function f become arbitrarily large as x → ∞ (or as x → −∞). In this case, we write
lim f (x) = ∞ (or
x→∞
lim f (x) = ∞). On the other hand, if the values of f are negative but become arbitrarily large in
x → −∞
magnitude as x → ∞ (or as x → −∞), we write x lim
f (x) = −∞ (or x →
lim
f (x) = −∞).
→∞
−∞
For example, consider the function f (x) = x 3. As seen in Table 4.9 and Figure 4.47, as x → ∞ the values f (x)
become arbitrarily large. Therefore, x lim
x 3 = ∞. On the other hand, as x → −∞,
→∞
the values of f (x) = x 3 are
lim
x 3 = −∞.
negative but become arbitrarily large in magnitude. Consequently, x →
−∞
x
10
20
50
100
1000
x3
1000
8000
125,000
1,000,000
1,000,000,000
x
−10
−20
−50
−100
−1000
x3
−1000
−8000
−125,000
−1,000,000
−1,000,000,000
Table 4.9 Values of a power function as x → ±∞
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Chapter 4 | Applications of Derivatives
413
Figure 4.47 For this function, the functional values approach infinity as
x → ±∞.
Definition
(Informal) We say a function f has an infinite limit at infinity and write
lim f (x) = ∞.
x→∞
if f (x) becomes arbitrarily large for x sufficiently large. We say a function has a negative infinite limit at infinity and
write
lim f (x) = −∞.
x→∞
if f (x) < 0 and | f (x)| becomes arbitrarily large for x sufficiently large. Similarly, we can define infinite limits as
x → −∞.
Formal Definitions
Earlier, we used the terms arbitrarily close, arbitrarily large, and sufficiently large to define limits at infinity informally.
Although these terms provide accurate descriptions of limits at infinity, they are not precise mathematically. Here are more
formal definitions of limits at infinity. We then look at how to use these definitions to prove results involving limits at
infinity.
Definition
(Formal) We say a function f has a limit at infinity, if there exists a real number L such that for all ε > 0, there
exists N > 0 such that
| f (x) − L| < ε
for all x > N. In that case, we write
lim f (x) = L
x→∞
(see Figure 4.48).
We say a function f has a limit at negative infinity if there exists a real number L such that for all ε > 0, there
exists N < 0 such that
| f (x) − L| < ε
for all x < N. In that case, we write
lim f (x) = L.
x → −∞
414
Chapter 4 | Applications of Derivatives
Figure 4.48 For a function with a limit at infinity, for all
x > N,
| f (x) − L| < ε.
Earlier in this section, we used graphical evidence in Figure 4.40 and numerical evidence in Table 4.8 to conclude that
⎛
⎞
lim 2 + 1 =
x → ∞⎝ x ⎠
2. Here we use the formal definition of limit at infinity to prove this result rigorously.
Example 4.22 A Finite Limit at Infinity Example
⎛
⎞
2 + 1 = 2.
Use the formal definition of limit at infinity to prove that x lim
→ ∞⎝ x ⎠
Solution
Let ε > 0. Let N = 1ε . Therefore, for all x > N, we have
|2 + 1x − 2| = |1x | = 1x < N1 = ε.
4.21
⎛3 − 1 ⎞
Use the formal definition of limit at infinity to prove that x lim
= 3.
→ ∞⎝ x 2 ⎠
We now turn our attention to a more precise definition for an infinite limit at infinity.
Definition
(Formal) We say a function f has an infinite limit at infinity and write
lim f (x) = ∞
x→∞
if for all M > 0, there exists an N > 0 such that
f (x) > M
for all x > N (see Figure 4.49).
We say a function has a negative infinite limit at infinity and write
lim f (x) = −∞
x→∞
if for all M < 0, there exists an N > 0 such that
f (x) < M
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Chapter 4 | Applications of Derivatives
415
for all x > N.
Similarly we can define limits as x → −∞.
Figure 4.49 For a function with an infinite limit at infinity, for
all x > N, f (x) > M.
Earlier, we used graphical evidence (Figure 4.47) and numerical evidence (Table 4.9) to conclude that x lim
x 3 = ∞.
→∞
Here we use the formal definition of infinite limit at infinity to prove that result.
Example 4.23 An Infinite Limit at Infinity
Use the formal definition of infinite limit at infinity to prove that x lim
x 3 = ∞.
→∞
Solution
3
Let M > 0. Let N = M . Then, for all x > N, we have
⎛3
⎞
3
x 3 > N 3 = ⎝ M ⎠ = M.
Therefore, x lim
x 3 = ∞.
→∞
4.22
Use the formal definition of infinite limit at infinity to prove that x lim
3x 2 = ∞.
→∞
End Behavior
The behavior of a function as x → ±∞ is called the function’s end behavior. At each of the function’s ends, the function
could exhibit one of the following types of behavior:
1. The function f (x) approaches a horizontal asymptote y = L.
2. The function f (x) → ∞ or f (x) → −∞.
3. The function does not approach a finite limit, nor does it approach ∞ or −∞. In this case, the function may have
some oscillatory behavior.
416
Chapter 4 | Applications of Derivatives
Let’s consider several classes of functions here and look at the different types of end behaviors for these functions.
End Behavior for Polynomial Functions
Consider the power function f (x) = x n where n is a positive integer. From Figure 4.50 and Figure 4.51, we see that
lim x n = ∞; n = 1, 2, 3,…
x→∞
and
⎧∞; n = 2, 4, 6,…
lim x n = ⎨
.
⎩−∞; n = 1, 3, 5,…
x → −∞
Figure 4.50 For power functions with an even power of n,
lim x n = ∞ = x →
lim
x n.
−∞
x→∞
Figure 4.51 For power functions with an odd power of n,
lim x n = ∞ and
x→∞
lim x n = −∞.
x → −∞
Using these facts, it is not difficult to evaluate x lim
cx n and x →
lim
cx n, where c is any constant and n is a positive
→∞
−∞
integer. If c > 0, the graph of y = cx n is a vertical stretch or compression of y = x n, and therefore
lim cx n = x lim
x n and x →
lim
cx n = x →
lim
x n if c > 0.
→∞
−∞
−∞
x→∞
If c < 0, the graph of y = cx n is a vertical stretch or compression combined with a reflection about the x -axis, and
therefore
lim cx n = −x lim
x n and x →
lim
cx n = −x →
lim
x n if c < 0.
→∞
−∞
−∞
x→∞
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Chapter 4 | Applications of Derivatives
417
cx n = 0 = x →
lim
cx n.
If c = 0, y = cx n = 0, in which case x lim
→∞
−∞
Example 4.24
Limits at Infinity for Power Functions
f (x) and x →
lim
f (x).
For each function f , evaluate x lim
→∞
−∞
a.
f (x) = −5x 3
b.
f (x) = 2x 4
Solution
a. Since the coefficient of x 3 is −5, the graph of f (x) = −5x 3 involves a vertical stretch and reflection
⎛
⎛
−5x 3⎞⎠ = −∞ and x →
−5x 3⎞⎠ = ∞.
of the graph of y = x 3 about the x -axis. Therefore, x lim
lim
→ ∞⎝
−∞⎝
b. Since the coefficient of x 4 is 2, the graph of f (x) = 2x 4 is a vertical stretch of the graph of y = x 4.
Therefore, x lim
2x 4 = ∞ and x →
lim
2x 4 = ∞.
→∞
−∞
4.23
Let f (x) = −3x 4. Find x lim
f (x).
→∞
We now look at how the limits at infinity for power functions can be used to determine
lim f (x) for any polynomial
x → ±∞
function f . Consider a polynomial function
f (x) = a n x n + a n − 1 x n − 1 + … + a 1 x + a 0
of degree n ≥ 1 so that a n ≠ 0. Factoring, we see that
a
⎛ a
f (x) = a n x n 1 + na− 1 1x + … + a 1
n
n
⎝
1 + a 0 ⎞.
an⎠
xn − 1
As x → ±∞, all the terms inside the parentheses approach zero except the first term. We conclude that
lim f (x) =
x → ±∞
lim a x
x → ±∞ n
n
.
For example, the function f (x) = 5x 3 − 3x 2 + 4 behaves like g(x) = 5x 3 as x → ±∞ as shown in Figure 4.52 and
Table 4.10.
418
Chapter 4 | Applications of Derivatives
Figure 4.52 The end behavior of a polynomial is determined
by the behavior of the term with the largest exponent.
10
100
1000
f(x) = 5x3 − 3x2 + 4
4704
4,970,004
4,997,000,004
g(x) = 5x3
5000
5,000,000
5,000,000,000
x
−10
−100
−1000
f(x) = 5x3 − 3x2 + 4
−5296
−5,029,996
−5,002,999,996
g(x) = 5x3
−5000
−5,000,000
−5,000,000,000
x
Table 4.10 A polynomial’s end behavior is determined by the term with the
largest exponent.
End Behavior for Algebraic Functions
The end behavior for rational functions and functions involving radicals is a little more complicated than for polynomials. In
Example 4.25, we show that the limits at infinity of a rational function f (x) =
p(x)
depend on the relationship between
q(x)
the degree of the numerator and the degree of the denominator. To evaluate the limits at infinity for a rational function,
we divide the numerator and denominator by the highest power of x appearing in the denominator. This determines which
term in the overall expression dominates the behavior of the function at large values of x.
Example 4.25
Determining End Behavior for Rational Functions
For each of the following functions, determine the limits as x → ∞ and x → −∞. Then, use this information
to describe the end behavior of the function.
a.
f (x) = 3x − 1 (Note: The degree of the numerator and the denominator are the same.)
2x + 5
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Chapter 4 | Applications of Derivatives
419
3x 2 + 2x (Note: The degree of numerator is less than the degree of the denominator.)
4x 3 − 5x + 7
b.
f (x) =
c.
2
f (x) = 3x + 4x (Note: The degree of numerator is greater than the degree of the denominator.)
x+2
Solution
a. The highest power of x in the denominator is x. Therefore, dividing the numerator and denominator by
x and applying the algebraic limit laws, we see that
lim 3x − 1 =
x → ±∞ 2x + 5
=
=
lim 3 − 1/x
x → ±∞ 2 + 5/x
lim (3 − 1/x)
x → ±∞
lim (2 + 5/x)
x → ±∞
lim 3 − lim 1/x
x → ±∞
x → ±∞
x → ±∞
x → ±∞
lim 2 + lim 5/x
= 3 − 0 = 3.
2+0 2
Since
lim f (x) = 3 , we know that y = 3 is a horizontal asymptote for this function as shown in
2
2
x → ±∞
the following graph.
Figure 4.53 The graph of this rational function approaches a
horizontal asymptote as x → ±∞.
b. Since the largest power of x appearing in the denominator is x 3, divide the numerator and denominator
by x 3. After doing so and applying algebraic limit laws, we obtain
lim
3x 2 + 2x
x → ±∞ 4x 3 − 5x + 7
=
lim
3/x + 2/x 2
x → ±∞ 4 − 5/x 2 + 7/x 3
=
3.0 + 2.0 = 0.
4 − 5.0 + 7.0
Therefore f has a horizontal asymptote of y = 0 as shown in the following graph.
420
Chapter 4 | Applications of Derivatives
Figure 4.54 The graph of this rational function approaches
the horizontal asymptote y = 0 as x → ±∞.
c. Dividing the numerator and denominator by x, we have
2
lim 3x + 4x = lim 3x + 4 .
x → ±∞ 1 + 2/x
x+2
x → ±∞
As x → ±∞,
the denominator approaches 1. As x → ∞, the numerator approaches +∞. As
x → −∞, the numerator approaches −∞. Therefore x lim
f (x) = ∞, whereas x →
lim
f (x) = −∞
→∞
−∞
as shown in the following figure.
Figure 4.55 As x → ∞, the values f (x) → ∞. As x → −∞, the
values f (x) → −∞.
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Chapter 4 | Applications of Derivatives
4.24
Evaluate
421
2
lim 3x + 2x − 1
x → ±∞ 5x 2 − 4x + 7
and use these limits to determine the end behavior of
2
f (x) = 3x 2 + 2x − 2 .
5x − 4x + 7
Before proceeding, consider the graph of f (x) =
⎛ 2
⎝3x
+ 4x⎞⎠
shown in Figure 4.56. As x → ∞ and x → −∞, the
(x + 2)
graph of f appears almost linear. Although f is certainly not a linear function, we now investigate why the graph of f
seems to be approaching a linear function. First, using long division of polynomials, we can write
2
f (x) = 3x + 4x = 3x − 2 + 4 .
x+2
x+2
Since
4 → 0 as x → ±∞, we conclude that
(x + 2)
lim
⎛
⎝
x → ±∞
f (x) − (3x − 2)⎞⎠ =
lim
4
x → ±∞ x + 2
= 0.
Therefore, the graph of f approaches the line y = 3x − 2 as x → ±∞. This line is known as an oblique asymptote for
f (Figure 4.56).
Figure 4.56 The graph of the rational function
f (x) = ⎛⎝3x 2 + 4x⎞⎠/(x + 2) approaches the oblique asymptote
y = 3x − 2 as x → ±∞.
We can summarize the results of Example 4.25 to make the following conclusion regarding end behavior for rational
functions. Consider a rational function
f (x) =
a xn + an − 1 xn − 1 + … + a1 x + a0
p(x)
= n m
,
m−1
q(x) b m x + b
+ … + b1 x + b0
m−1x
where a n ≠ 0 and b m ≠ 0.
1. If the degree of the numerator is the same as the degree of the denominator (n = m), then f has a horizontal
asymptote of y = a n /b m as x → ±∞.
2. If the degree of the numerator is less than the degree of the denominator (n < m), then f has a horizontal
asymptote of y = 0 as x → ±∞.
3. If the degree of the numerator is greater than the degree of the denominator (n > m), then f does not have a
422
Chapter 4 | Applications of Derivatives
horizontal asymptote. The limits at infinity are either positive or negative infinity, depending on the signs of the
leading terms. In addition, using long division, the function can be rewritten as
f (x) =
p(x)
r(x)
= g(x) +
,
q(x)
q(x)
where the degree of r(x) is less than the degree of q(x). As a result,
⎡
⎣
⎤
⎦
lim r(x)/q(x) = 0. Therefore, the values
x → ±∞
of f (x) − g(x) approach zero as x → ±∞. If the degree of p(x) is exactly one more than the degree of q(x)
(n = m + 1), the function g(x) is a linear function. In this case, we call g(x) an oblique asymptote.
Now let’s consider the end behavior for functions involving a radical.
Example 4.26
Determining End Behavior for a Function Involving a Radical
Find the limits as x → ∞ and x → −∞ for f (x) = 3x − 2
4x 2 + 5
and describe the end behavior of f .
Solution
Let’s use the same strategy as we did for rational functions: divide the numerator and denominator by a power of
x. To determine the appropriate power of x, consider the expression 4x 2 + 5 in the denominator. Since
4x 2 + 5 ≈ 4x 2 = 2|x|
for large values of x in effect x appears just to the first power in the denominator. Therefore, we divide the
numerator and denominator by |x|. Then, using the fact that |x| = x for x > 0, |x| = −x for x < 0, and
|x| = x 2 for all x, we calculate the limits as follows:
lim
x→∞
lim
x → −∞
3x − 2 =
4x 2 + 5
lim
x→∞
(1/|x|)(3x − 2)
(1/|x|) 4x 2 + 5
(1/x)(3x − 2)
=
x→∞ ⎛
=
x→∞
3x − 2 =
4x 2 + 5
lim
2⎞⎛ 2
⎝1/x ⎠⎝4x
lim
lim
x → −∞
+ 5⎞⎠
3 − 2/x = 3 = 3
4 2
4 + 5/x 2
(1/|x|)(3x − 2)
(1/|x|) 4x 2 + 5
(−1/x)(3x − 2)
=
x → −∞ ⎛
lim
=
x → −∞
2⎞⎛ 2
⎝1/x ⎠⎝4x
+ 5⎞⎠
lim −3 + 2/x = −3 = −3 .
2
4
4 + 5/x 2
Therefore, f (x) approaches the horizontal asymptote y = 3 as x → ∞ and the horizontal asymptote y = − 3
as x → −∞ as shown in the following graph.
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2
2
Chapter 4 | Applications of Derivatives
423
Figure 4.57 This function has two horizontal asymptotes and it crosses one
of the asymptotes.
4.25
3x 2 + 4 .
Evaluate x lim
→∞ x+6
Determining End Behavior for Transcendental Functions
The six basic trigonometric functions are periodic and do not approach a finite limit as x → ±∞. For example, sin x
oscillates between 1 and −1 (Figure 4.58). The tangent function x has an infinite number of vertical asymptotes as
x → ±∞; therefore, it does not approach a finite limit nor does it approach ±∞ as x → ±∞ as shown in Figure 4.59.
Figure 4.58 The function f (x) = sin x oscillates between
1 and −1 as x → ±∞
Figure 4.59 The function f (x) = tan x does not approach a
limit and does not approach ±∞ as x → ±∞
Recall that for any base b > 0, b ≠ 1, the function y = b x is an exponential function with domain (−∞, ∞) and range
(0, ∞). If b > 1, y = b x is increasing over `(−∞, ∞). If 0 < b < 1, y = b x is decreasing over (−∞, ∞). For
the natural exponential function f (x) = e x, e ≈ 2.718 > 1. Therefore, f (x) = e x is increasing on `(−∞, ∞) and the
424
Chapter 4 | Applications of Derivatives
range is `(0, ∞). The exponential function f (x) = e x approaches ∞ as x → ∞ and approaches 0 as x → −∞ as
shown in Table 4.11 and Figure 4.60.
x
−5
−2
0
2
5
ex
0.00674
0.135
1
7.389
148.413
Table 4.11 End behavior of the natural exponential function
Figure 4.60 The exponential function approaches zero as
x → −∞ and approaches ∞ as x → ∞.
Recall that the natural logarithm function f (x) = ln(x) is the inverse of the natural exponential function y = e x.
Therefore, the domain of f (x) = ln(x) is (0, ∞) and the range is (−∞, ∞). The graph of f (x) = ln(x) is the reflection
of the graph of y = e x about the line y = x. Therefore, ln(x) → −∞ as x → 0+ and ln(x) → ∞ as x → ∞ as shown
in Figure 4.61 and Table 4.12.
x
ln(x)
0.01
0.1
1
10
100
−4.605
−2.303
0
2.303
4.605
Table 4.12 End behavior of the natural logarithm function
Figure 4.61 The natural logarithm function approaches ∞ as
x → ∞.
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Chapter 4 | Applications of Derivatives
425
Example 4.27
Determining End Behavior for a Transcendental Function
Find the limits as x → ∞ and x → −∞ for f (x) =
(2 + 3e x)
and describe the end behavior of f .
(7 − 5e x)
Solution
To find the limit as x → ∞, divide the numerator and denominator by e x :
x
2 + 3e
lim f (x) = x lim
→ ∞ 7 − 5e x
(2/e x) + 3
= x lim
.
→ ∞ (7/e x) − 5
x→∞
As shown in Figure 4.60, e x → ∞ as x → ∞. Therefore,
7.
lim 2 = 0 = x lim
→ ∞e x
x → ∞e x
We conclude that x lim
f (x) = − 3 , and the graph of f approaches the horizontal asymptote y = − 3
→∞
5
5
as x → ∞. To find the limit as x → −∞,
use the fact that e x → 0 as x → −∞ to conclude that
lim f (x) = 2 , and therefore the graph of approaches the horizontal asymptote y = 2 as x → −∞.
7
7
x→∞
4.26
Find the limits as x → ∞ and x → −∞ for f (x) =
(3e x − 4)
.
(5e x + 2)
Guidelines for Drawing the Graph of a Function
We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before
showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.
Problem-Solving Strategy: Drawing the Graph of a Function
Given a function f , use the following steps to sketch a graph of f :
1. Determine the domain of the function.
2. Locate the x - and y -intercepts.
3. Evaluate x lim
f (x) and x →
lim
f (x) to determine the end behavior. If either of these limits is a finite number
→∞
−∞
L, then y = L is a horizontal asymptote. If either of these limits is ∞ or −∞, determine whether f has
an oblique asymptote. If f is a rational function such that f (x) =
p(x)
, where the degree of the numerator
q(x)
is greater than the degree of the denominator, then f can be written as
f (x) =
p(x)
r(x)
= g(x) +
,
q(x)
q(x)
where the degree of r(x) is less than the degree of q(x). The values of f (x) approach the values of g(x) as
426
Chapter 4 | Applications of Derivatives
x → ±∞. If g(x) is a linear function, it is known as an oblique asymptote.
4. Determine whether f has any vertical asymptotes.
5. Calculate f ′. Find all critical points and determine the intervals where f is increasing and where f is
decreasing. Determine whether f has any local extrema.
6. Calculate f ″. Determine the intervals where f is concave up and where f is concave down. Use this
information to determine whether f has any inflection points. The second derivative can also be used as an
alternate means to determine or verify that f has a local extremum at a critical point.
Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.
Example 4.28
Sketching a Graph of a Polynomial
Sketch a graph of f (x) = (x − 1) 2 (x + 2).
Solution
Step 1. Since f is a polynomial, the domain is the set of all real numbers.
Step 2. When x = 0, f (x) = 2. Therefore, the y -intercept is (0, 2). To find the x -intercepts, we need to solve
the equation (x − 1) 2 (x + 2) = 0, gives us the x -intercepts (1, 0) and (−2, 0)
Step 3. We need to evaluate the end behavior of f . As x → ∞,
lim f (x) = ∞. As x → −∞,
x→∞
(x − 1) 2 → ∞ and (x + 2) → ∞. Therefore,
(x − 1) 2 → ∞ and (x + 2) → −∞. Therefore,
lim f (x) = −∞. To get
x→∞
even more information about the end behavior of f , we can multiply the factors of f . When doing so, we see
that
f (x) = (x − 1) 2 (x + 2) = x 3 − 3x + 2.
Since the leading term of f is x 3, we conclude that f behaves like y = x 3 as x → ±∞.
Step 4. Since f is a polynomial function, it does not have any vertical asymptotes.
Step 5. The first derivative of f is
f ′ (x) = 3x 2 − 3.
Therefore, f has two critical points: x = 1, −1. Divide the interval (−∞, ∞) into the three smaller intervals:
(−∞, −1),
(−1, 1), and (1, ∞). Then, choose test points x = −2,
x = 0, and x = 2 from these
intervals and evaluate the sign of f ′ (x) at each of these test points, as shown in the following table.
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Chapter 4 | Applications of Derivatives
427
Interval
Test
Point
Sign of Derivative
f′(x) = 3x2 − 3 = 3(x − 1)(x + 1)
(−∞, −1)
x = −2
(+)(−)(−) = +
Conclusion
f is
increasing.
(−1, 1)
x=0
(+)(−)(+) = −
f is
decreasing.
(1, ∞)
x=2
(+)(+)(+) = +
f is
increasing.
From the table, we see that f has a local maximum at x = −1 and a local minimum at x = 1. Evaluating
f (x) at those two points, we find that the local maximum value is f (−1) = 4 and the local minimum value is
f (1) = 0.
Step 6. The second derivative of f is
f ″(x) = 6x.
The second derivative is zero at x = 0. Therefore, to determine the concavity of f ,
divide the interval
(−∞, ∞) into the smaller intervals (−∞, 0) and (0, ∞), and choose test points x = −1 and x = 1 to
determine the concavity of f on each of these smaller intervals as shown in the following table.
Sign of f″(x) = 6x
Conclusion
Interval
Test Point
(−∞, 0)
x = −1
−
f is concave down.
(0, ∞)
x=1
+
f is concave up.
We note that the information in the preceding table confirms the fact, found in step 5, that f has a local
maximum at x = −1 and a local minimum at x = 1. In addition, the information found in step 5 —namely, f
has a local maximum at x = −1 and a local minimum at x = 1, and f ′ (x) = 0 at those points—combined
with the fact that f ″ changes sign only at x = 0 confirms the results found in step 6 on the concavity of f .
Combining this information, we arrive at the graph of f (x) = (x − 1) 2 (x + 2) shown in the following graph.
428
Chapter 4 | Applications of Derivatives
4.27
Sketch a graph of f (x) = (x − 1) 3 (x + 2).
Example 4.29
Sketching a Rational Function
Sketch the graph of f (x) =
⎛
⎝1
x2 .
− x 2⎞⎠
Solution
Step 1. The function f is defined as long as the denominator is not zero. Therefore, the domain is the set of all
real numbers x except x = ±1.
Step 2. Find the intercepts. If x = 0, then f (x) = 0, so 0 is an intercept. If y = 0, then
⎛
⎝1
x 2 = 0,
− x 2⎞⎠
which implies x = 0. Therefore, (0, 0) is the only intercept.
Step 3. Evaluate the limits at infinity. Since f is a rational function, divide the numerator and denominator by
the highest power in the denominator: x 2. We obtain
lim
x2
x → ±∞ 1 − x 2
=
lim
x → ±∞ 1
x2
1 = −1.
−1
Therefore, f has a horizontal asymptote of y = −1 as x → ∞ and x → −∞.
Step 4. To determine whether f has any vertical asymptotes, first check to see whether the denominator has any
zeroes. We find the denominator is zero when x = ±1. To determine whether the lines x = 1 or x = −1 are
vertical asymptotes of f , evaluate lim f (x) and
x→1
lim f (x). By looking at each one-sided limit as x → 1,
x → −1
we see that
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Chapter 4 | Applications of Derivatives
429
lim
x → 1+
x 2 = −∞ and lim
x 2 = ∞.
−
2
x
→
1
1−x
1 − x2
In addition, by looking at each one-sided limit as x → −1, we find that
lim
x → −1+
x 2 = ∞ and lim
x 2 = −∞.
−
2
x → −1 1 − x 2
1−x
Step 5. Calculate the first derivative:
f ′ (x) =
⎛
⎝1
− x 2⎞⎠(2x) − x 2 (−2x)
⎛
⎝1
−
x 2⎞⎠
2
=
⎛
⎝1
2x
−x
2⎞
⎠
2
.
Critical points occur at points x where f ′ (x) = 0 or f ′ (x) is undefined. We see that f ′ (x) = 0 when x = 0.
The derivative f ′ is not undefined at any point in the domain of f . However, x = ±1 are not in the domain of
f . Therefore, to determine where f is increasing and where f is decreasing, divide the interval (−∞, ∞) into
four smaller intervals: (−∞, −1),
(−1, 0), (0, 1), and (1, ∞), and choose a test point in each interval to
determine the sign of f ′ (x) in each of these intervals. The values x = −2, x = − 1 , x = 1 , and x = 2
2
2
are good choices for test points as shown in the following table.
Interval
Test Point
Sign of f′ (x) =
2x
⎛
⎝1
−x
2
2⎞
Conclusion
⎠
(−∞, −1)
x = −2
−/ + = −
f is decreasing.
(−1, 0)
x = −1/2
−/ + = −
f is decreasing.
(0, 1)
x = 1/2
+/ + = +
f is increasing.
(1, ∞)
x=2
+/ + = +
f is increasing.
From this analysis, we conclude that f has a local minimum at x = 0 but no local maximum.
Step 6. Calculate the second derivative:
430
Chapter 4 | Applications of Derivatives
f ″(x) =
=
=
⎛
⎝1
⎛
⎝1
2
− x 2⎞⎠ (2) − 2x⎛⎝2⎛⎝1 − x 2⎞⎠(−2x)⎞⎠
⎛
⎝1
− x 2⎞⎠
4
− x 2⎞⎠⎡⎣2⎛⎝1 − x 2⎞⎠ + 8x 2⎤⎦
⎛
⎝1
− x 2⎞⎠
4
2⎛⎝1 − x 2⎞⎠ + 8x 2
⎛
⎝1
− x 2⎞⎠
3
2
= 6x + 23 .
⎛
2⎞
⎝1 − x ⎠
To determine the intervals where f is concave up and where f is concave down, we first need to find all points
x where f ″(x) = 0 or f ″(x) is undefined. Since the numerator 6x 2 + 2 ≠ 0 for any x,
f ″(x) is never zero.
Furthermore, f ″ is not undefined for any x in the domain of f . However, as discussed earlier, x = ±1 are
not in the domain of f . Therefore, to determine the concavity of f , we divide the interval (−∞, ∞) into the
three smaller intervals (−∞, −1),
(−1, −1), and (1, ∞), and choose a test point in each of these intervals
to evaluate the sign of f ″(x). in each of these intervals. The values x = −2,
x = 0, and x = 2 are possible
test points as shown in the following table.
Interval
Test Point
2
Sign of f″(x) = 6x + 23
⎛
2⎞
⎝1 − x ⎠
(−∞, −1)
x = −2
+/ − = −
f is concave down.
(−1, −1)
x=0
+/ + = +
f is concave up.
(1, ∞)
x=2
+/ − = −
f is concave down.
Conclusion
Combining all this information, we arrive at the graph of f shown below. Note that, although f changes
concavity at x = −1 and x = 1, there are no inflection points at either of these places because f is not
continuous at x = −1 or x = 1.
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Chapter 4 | Applications of Derivatives
4.28
431
Sketch a graph of f (x) =
(3x + 5)
.
(8 + 4x)
Example 4.30
Sketching a Rational Function with an Oblique Asymptote
Sketch the graph of f (x) =
x2
(x − 1)
Solution
Step 1. The domain of f is the set of all real numbers x except x = 1.
Step 2. Find the intercepts. We can see that when x = 0,
f (x) = 0, so (0, 0) is the only intercept.
Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the
denominator, f must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials
to write
2
f (x) = x = x + 1 + 1 .
x−1
x−1
Since 1/(x − 1) → 0 as x → ±∞,
f (x) approaches the line y = x + 1 as x → ±∞. The line y = x + 1 is
an oblique asymptote for f .
Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at
x = 1. Looking at both one-sided limits as x → 1, we find
lim
x → 1+
x 2 = ∞ and lim x 2 = −∞.
x−1
x → 1− x − 1
Therefore, x = 1 is a vertical asymptote, and we have determined the behavior of f as x approaches 1 from
the right and the left.
432
Chapter 4 | Applications of Derivatives
Step 5. Calculate the first derivative:
f ′ (x) =
(x − 1)(2x) − x 2 (1) x 2 − 2x
=
.
(x − 1) 2
(x − 1) 2
We have f ′ (x) = 0 when x 2 − 2x = x(x − 2) = 0. Therefore, x = 0 and x = 2 are critical points. Since f
is undefined at x = 1, we need to divide the interval (−∞, ∞) into the smaller intervals (−∞, 0),
(0, 1),
(1, 2), and (2, ∞), and choose a test point from each interval to evaluate the sign of f ′ (x) in each of these
smaller intervals. For example, let x = −1,
x = 1,
2
x = 3 , and x = 3 be the test points as shown in the
2
following table.
Interval
Test Point
2
x(x − 2)
Sign of f′(x) = x − 2x2 =
(x − 1) 2
(x − 1)
(−∞, 0)
x = −1
(−)(−)/ + = +
f is increasing.
(0, 1)
x = 1/2
(+)(−)/ + = −
f is decreasing.
(1, 2)
x = 3/2
(+)(−)/ + = −
f is decreasing.
(2, ∞)
x=3
(+)(+)/ + = +
f is increasing.
Conclusion
From this table, we see that f has a local maximum at x = 0 and a local minimum at x = 2. The value of f
at the local maximum is f (0) = 0 and the value of f at the local minimum is f (2) = 4. Therefore, (0, 0) and
(2, 4) are important points on the graph.
Step 6. Calculate the second derivative:
f ″(x) =
=
=
=
=
(x − 1) 2 (2x − 2) − ⎛⎝x 2 − 2x⎞⎠⎛⎝2(x − 1)⎞⎠
(x − 1) 4
(x − 1)⎡⎣(x − 1)(2x − 2) − 2⎛⎝x 2 − 2x⎞⎠⎤⎦
(x − 1) 4
(x − 1)(2x − 2) − 2⎛⎝x 2 − 2x⎞⎠
(x − 1) 3
2x 2 − 4x + 2 − ⎛⎝2x 2 − 4x⎞⎠
2 .
(x − 1) 3
(x − 1) 3
We see that f ″(x) is never zero or undefined for x in the domain of f . Since f is undefined at x = 1, to
check concavity we just divide the interval (−∞, ∞) into the two smaller intervals (−∞, 1) and (1, ∞), and
choose a test point from each interval to evaluate the sign of f ″(x) in each of these intervals. The values x = 0
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Chapter 4 | Applications of Derivatives
433
and x = 2 are possible test points as shown in the following table.
2
(x − 1) 3
Interval
Test Point
Sign of f″(x) =
Conclusion
(−∞, 1)
x=0
+/ − = −
f is concave down.
(1, ∞)
x=2
+/ + = +
f is concave up.
From the information gathered, we arrive at the following graph for f .
4.29
Find the oblique asymptote for f (x) =
⎛ 3
⎝3x − 2x +
⎛ 2
⎞
⎝2x − 4⎠
1⎞⎠
.
Example 4.31
Sketching the Graph of a Function with a Cusp
Sketch a graph of f (x) = (x − 1) 2/3.
Solution
⎛3
⎞
Step 1. Since the cube-root function is defined for all real numbers x and (x − 1) 2/3 = ⎝ x − 1⎠ , the domain
2
of f is all real numbers.
Step 2: To find the y -intercept, evaluate f (0). Since f (0) = 1, the y -intercept is (0, 1). To find the x
-intercept, solve (x − 1) 2/3 = 0. The solution of this equation is x = 1, so the x -intercept is (1, 0).
434
Chapter 4 | Applications of Derivatives
Step 3: Since
lim (x − 1) 2/3 = ∞, the function continues to grow without bound as x → ∞ and x → −∞.
x → ±∞
Step 4: The function has no vertical asymptotes.
Step 5: To determine where f is increasing or decreasing, calculate f ′. We find
2
f ′ (x) = 2 (x − 1) −1/3 =
.
3
3(x − 1) 1/3
This function is not zero anywhere, but it is undefined when x = 1. Therefore, the only critical point is x = 1.
Divide the interval (−∞, ∞) into the smaller intervals (−∞, 1) and (1, ∞), and choose test points in each
of these intervals to determine the sign of f ′ (x) in each of these smaller intervals. Let x = 0 and x = 2 be the
test points as shown in the following table.
2
3(x − 1) 1/3
Conclusion
Interval
Test Point
Sign of f′ (x) =
(−∞, 1)
x=0
+/ − = −
f is decreasing.
(1, ∞)
x=2
+/ + = +
f is increasing.
We conclude that f has a local minimum at x = 1. Evaluating f at x = 1, we find that the value of f at the
local minimum is zero. Note that f ′ (1) is undefined, so to determine the behavior of the function at this critical
point, we need to examine lim f ′ (x). Looking at the one-sided limits, we have
x→1
lim
x → 1+
2
2
= ∞ and lim−
= −∞.
x → 1 3(x − 1) 1/3
3(x − 1) 1/3
Therefore, f has a cusp at x = 1.
Step 6: To determine concavity, we calculate the second derivative of f :
−2
f ″(x) = − 2 (x − 1) −4/3 =
.
9
9(x − 1) 4/3
We find that f ″(x) is defined for all x, but is undefined when x = 1. Therefore, divide the interval (−∞, ∞)
into the smaller intervals (−∞, 1) and (1, ∞), and choose test points to evaluate the sign of f ″(x) in each of
these intervals. As we did earlier, let x = 0 and x = 2 be test points as shown in the following table.
−2
9(x − 1) 4/3
Interval
Test Point
Sign of f″(x) =
(−∞, 1)
x=0
−/ + = −
f is concave down.
(1, ∞)
x=2
−/ + = −
f is concave down.
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Conclusion
Chapter 4 | Applications of Derivatives
435
From this table, we conclude that f is concave down everywhere. Combining all of this information, we arrive
at the following graph for f .
4.30
Consider the function f (x) = 5 − x 2/3. Determine the point on the graph where a cusp is located.
Determine the end behavior of f .
436
Chapter 4 | Applications of Derivatives
4.6 EXERCISES
For the following exercises, examine the graphs. Identify
where the vertical asymptotes are located.
254.
251.
255.
252.
For the following functions f (x),
253.
determine whether
there is an asymptote at x = a. Justify your answer
without graphing on a calculator.
256.
f (x) =
x + 1 , a = −1
x 2 + 5x + 4
257.
f (x) =
x ,a=2
x−2
258.
f (x) = (x + 2) 3/2, a = −2
259.
f (x) = (x − 1) −1/3, a = 1
260.
f (x) = 1 + x −2/5, a = 1
For the following exercises, evaluate the limit.
1
261. x lim
→ ∞ 3x + 6
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Chapter 4 | Applications of Derivatives
437
2x − 5
262. x lim
→ ∞ 4x
281.
3
f (x) = x 3 + 1
x −1
x 2 − 2x + 5
263. x lim
→∞
x+2
282.
f (x) = sin x + cos x
sin x − cos x
283.
f (x) = x − sin x
284.
f (x) = 1x − x
3
3x − 2x
264. x →
lim
−∞ x 2 + 2x + 8
x 4 − 4x 3 + 1
265. x →
lim
−∞ 2 − 2x 2 − 7x 4
266. x lim
→∞
3x
2
x +1
4x 2 − 1
267. x →
lim
−∞ x + 2
268. x lim
→∞
269. x →
lim
−∞
4x
x2 − 1
4x
x2 − 1
For the following exercises, construct a function f (x) that
has the given asymptotes.
285. x = 1 and y = 2
286. x = 1 and y = 0
287. y = 4,
x = −1
288. x = 0
For the following exercises, graph the function on a
graphing calculator on the window x = ⎡⎣−5, 5⎤⎦ and
2 x
270. x lim
→ ∞x − x + 1
estimate the horizontal asymptote or limit. Then, calculate
the actual horizontal asymptote or limit.
For the following exercises, find the horizontal and vertical
asymptotes.
289. [T] f (x) =
271.
f (x) = x − 9x
272.
f (x) =
1
1 − x2
273.
f (x) =
x3
4 − x2
274.
2
f (x) = x 2 + 3
x +1
275.
f (x) = sin(x)sin(2x)
276.
f (x) = cos x + cos(3x) + cos(5x)
277.
x sin(x)
f (x) = 2
x −1
278.
f (x) =
x
sin(x)
279.
f (x) =
1
x3 + x2
f (x) =
1 − 2x
x−1
280.
1
x + 10
290. [T] f (x) = 2 x + 1
x + 7x + 6
291. [T] x →
lim
x 2 + 10x + 25
−∞
x+2
292. [T] x →
lim
−∞ x 2 + 7x + 6
3x + 2
293. [T] x lim
→∞ x+5
For the following exercises, draw a graph of the functions
without using a calculator. Be sure to notice all important
features of the graph: local maxima and minima, inflection
points, and asymptotic behavior.
294. y = 3x 2 + 2x + 4
295. y = x 3 − 3x 2 + 4
296. y = 2 2x + 1
x + 6x + 5
3
2
297. y = x + 4x + 3x
3x + 9
438
Chapter 4 | Applications of Derivatives
2
298. y = x2 + x − 2
x − 3x − 4
299. y = x 2 − 5x + 4
300. y = 2x 16 − x 2
x
301. y = cos
x , on x = [−2π, 2π]
302. y = e x − x 3
303. y = x tan x, x = [−π, π]
304. y = x ln(x), x > 0
305. y = x 2 sin(x), x = [−2π, 2π]
306. For f (x) =
P(x)
to have an asymptote at y = 2
Q(x)
then the polynomials P(x) and Q(x) must have what
relation?
307. For f (x) =
P(x)
to have an asymptote at x = 0,
Q(x)
then the polynomials P(x) and Q(x). must have what
relation?
308. If f ′ (x) has asymptotes at y = 3 and x = 1, then
f (x) has what asymptotes?
309.
Both f (x) =
1
1
and g(x) =
have
(x − 1)
(x − 1) 2
asymptotes at x = 1 and y = 0. What is the most obvious
difference between these two functions?
310. True or false: Every ratio of polynomials has vertical
asymptotes.
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Chapter 4 | Applications of Derivatives
439
4.7 | Applied Optimization Problems
Learning Objectives
4.7.1 Set up and solve optimization problems in several applied fields.
One common application of calculus is calculating the minimum or maximum value of a function. For example, companies
often want to minimize production costs or maximize revenue. In manufacturing, it is often desirable to minimize the
amount of material used to package a product with a certain volume. In this section, we show how to set up these types of
minimization and maximization problems and solve them by using the tools developed in this chapter.
Solving Optimization Problems over a Closed, Bounded Interval
The basic idea of the optimization problems that follow is the same. We have a particular quantity that we are interested
in maximizing or minimizing. However, we also have some auxiliary condition that needs to be satisfied. For example, in
Example 4.32, we are interested in maximizing the area of a rectangular garden. Certainly, if we keep making the side
lengths of the garden larger, the area will continue to become larger. However, what if we have some restriction on how
much fencing we can use for the perimeter? In this case, we cannot make the garden as large as we like. Let’s look at how
we can maximize the area of a rectangle subject to some constraint on the perimeter.
Example 4.32
Maximizing the Area of a Garden
A rectangular garden is to be constructed using a rock wall as one side of the garden and wire fencing for the
other three sides (Figure 4.62). Given 100 ft of wire fencing, determine the dimensions that would create a
garden of maximum area. What is the maximum area?
Figure 4.62 We want to determine the measurements x and
y that will create a garden with a maximum area using 100 ft
of fencing.
Solution
Let x denote the length of the side of the garden perpendicular to the rock wall and y denote the length of the
side parallel to the rock wall. Then the area of the garden is
A = x · y.
440
Chapter 4 | Applications of Derivatives
We want to find the maximum possible area subject to the constraint that the total fencing is 100 ft. From Figure
4.62, the total amount of fencing used will be 2x + y. Therefore, the constraint equation is
2x + y = 100.
Solving this equation for y, we have y = 100 − 2x. Thus, we can write the area as
A(x) = x · (100 − 2x) = 100x − 2x 2.
Before trying to maximize the area function A(x) = 100x − 2x 2, we need to determine the domain under
consideration. To construct a rectangular garden, we certainly need the lengths of both sides to be positive.
Therefore, we need x > 0 and y > 0. Since y = 100 − 2x, if y > 0, then x < 50. Therefore, we are trying
to determine the maximum value of A(x) for x over the open interval (0, 50). We do not know that a function
necessarily has a maximum value over an open interval. However, we do know that a continuous function has
an absolute maximum (and absolute minimum) over a closed interval. Therefore, let’s consider the function
A(x) = 100x − 2x 2 over the closed interval ⎡⎣0, 50⎤⎦. If the maximum value occurs at an interior point, then
we have found the value x in the open interval (0, 50) that maximizes the area of the garden. Therefore, we
consider the following problem:
Maximize A(x) = 100x − 2x 2 over the interval ⎡⎣0, 50⎤⎦.
As mentioned earlier, since A is a continuous function on a closed, bounded interval, by the extreme value
theorem, it has a maximum and a minimum. These extreme values occur either at endpoints or critical points. At
the endpoints, A(x) = 0. Since the area is positive for all x in the open interval (0, 50), the maximum must
occur at a critical point. Differentiating the function A(x), we obtain
A′ (x) = 100 − 4x.
Therefore, the only critical point is x = 25 (Figure 4.63). We conclude that the maximum area must occur when
x = 25. Then we have y = 100 − 2x = 100 − 2(25) = 50. To maximize the area of the garden, let x = 25 ft
and y = 50 ft. The area of this garden is 1250 ft 2.
Figure 4.63 To maximize the area of the garden, we need to find the
maximum value of the function A(x) = 100x − 2x 2.
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Chapter 4 | Applications of Derivatives
441
4.31 Determine the maximum area if we want to make the same rectangular garden as in Figure 4.63, but
we have 200 ft of fencing.
Now let’s look at a general strategy for solving optimization problems similar to Example 4.32.
Problem-Solving Strategy: Solving Optimization Problems
1. Introduce all variables. If applicable, draw a figure and label all variables.
2. Determine which quantity is to be maximized or minimized, and for what range of values of the other variables
(if this can be determined at this time).
3. Write a formula for the quantity to be maximized or minimized in terms of the variables. This formula may
involve more than one variable.
4. Write any equations relating the independent variables in the formula from step 3. Use these equations to
write the quantity to be maximized or minimized as a function of one variable.
5. Identify the domain of consideration for the function in step 4 based on the physical problem to be solved.
6. Locate the maximum or minimum value of the function from step 4. This step typically involves looking for
critical points and evaluating a function at endpoints.
Now let’s apply this strategy to maximize the volume of an open-top box given a constraint on the amount of material to be
used.
Example 4.33
Maximizing the Volume of a Box
An open-top box is to be made from a 24 in. by 36 in. piece of cardboard by removing a square from each
corner of the box and folding up the flaps on each side. What size square should be cut out of each corner to get
a box with the maximum volume?
Solution
Step 1: Let x be the side length of the square to be removed from each corner (Figure 4.64). Then, the
remaining four flaps can be folded up to form an open-top box. Let V be the volume of the resulting box.
442
Chapter 4 | Applications of Derivatives
Figure 4.64 A square with side length x inches is removed from each
corner of the piece of cardboard. The remaining flaps are folded to form an
open-top box.
Step 2: We are trying to maximize the volume of a box. Therefore, the problem is to maximize V.
Step 3: As mentioned in step 2,
are trying to maximize the volume of a box. The volume of a box is
V = L · W · H, where L, W, and H are the length, width, and height, respectively.
Step 4: From Figure 4.64, we see that the height of the box is x inches, the length is 36 − 2x inches, and the
width is 24 − 2x inches. Therefore, the volume of the box is
V(x) = (36 − 2x)(24 − 2x)x = 4x 3 − 120x 2 + 864x.
Step 5: To determine the domain of consideration, let’s examine Figure 4.64. Certainly, we need x > 0.
Furthermore, the side length of the square cannot be greater than or equal to half the length of the shorter side, 24
in.; otherwise, one of the flaps would be completely cut off. Therefore, we are trying to determine whether there
is a maximum volume of the box for x over the open interval (0, 12). Since V is a continuous function over
the closed interval [0, 12], we know V will have an absolute maximum over the closed interval. Therefore,
we consider V over the closed interval [0, 12] and check whether the absolute maximum occurs at an interior
point.
Step 6: Since V(x) is a continuous function over the closed, bounded interval [0, 12], V must have an absolute
maximum (and an absolute minimum). Since V(x) = 0 at the endpoints and V(x) > 0 for 0 < x < 12, the
maximum must occur at a critical point. The derivative is
V′ (x) = 12x 2 − 240x + 864.
To find the critical points, we need to solve the equation
12x 2 − 240x + 864 = 0.
Dividing both sides of this equation by 12, the problem simplifies to solving the equation
x 2 − 20x + 72 = 0.
Using the quadratic formula, we find that the critical points are
x=
20± (−20) 2 − 4(1)(72) 20± 112 20±4 7
=
=
= 10±2 7.
2
2
2
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Chapter 4 | Applications of Derivatives
443
Since 10 + 2 7 is not in the domain of consideration, the only critical point we need to consider is 10 − 2 7.
Therefore,
the
volume
is
maximized
if
3
we
let
x = 10 − 2 7 in.
The
maximum
volume
is
V(10 − 2 7) = 640 + 448 7 ≈ 1825 in. as shown in the following graph.
Figure 4.65 Maximizing the volume of the box leads to finding the maximum value of a
cubic polynomial.
Watch a video (http://www.openstaxcollege.org/l/20_boxvolume) about optimizing the volume of a box.
4.32 Suppose the dimensions of the cardboard in Example 4.33 are 20 in. by 30 in. Let x be the side length
of each square and write the volume of the open-top box as a function of x. Determine the domain of
consideration for x.
Example 4.34
Minimizing Travel Time
An island is 2 mi due north of its closest point along a straight shoreline. A visitor is staying at a cabin on the
shore that is 6 mi west of that point. The visitor is planning to go from the cabin to the island. Suppose the
visitor runs at a rate of 8 mph and swims at a rate of 3 mph. How far should the visitor run before swimming
to minimize the time it takes to reach the island?
Solution
Step 1: Let x be the distance running and let y be the distance swimming (Figure 4.66). Let T be the time it
takes to get from the cabin to the island.
444
Chapter 4 | Applications of Derivatives
Figure 4.66 How can we choose x and y to minimize the travel time from
the cabin to the island?
Step 2: The problem is to minimize T.
Step 3: To find the time spent traveling from the cabin to the island, add the time spent running and the time spent
swimming. Since Distance = Rate × Time (D = R × T), the time spent running is
T running =
D running x
= ,
R running 8
and the time spent swimming is
T swimming =
D swimming y
= .
R swimming 3
Therefore, the total time spent traveling is
y
T=x+ .
8 3
Step 4: From Figure 4.66, the line segment of y miles forms the hypotenuse of a right triangle with legs
of length 2 mi and 6 − x mi. Therefore, by the Pythagorean theorem, 2 2 + (6 − x) 2 = y 2, and we obtain
y = (6 − x) 2 + 4. Thus, the total time spent traveling is given by the function
(6 − x) 2 + 4
T(x) = x +
.
8
3
Step 5: From Figure 4.66, we see that 0 ≤ x ≤ 6. Therefore, ⎡⎣0, 6⎤⎦ is the domain of consideration.
Step 6: Since T(x) is a continuous function over a closed, bounded interval, it has a maximum and a minimum.
Let’s begin by looking for any critical points of T over the interval ⎡⎣0, 6⎤⎦. The derivative is
⎡
⎤
2
⎣(6 − x) + 4⎦
T′ (x) = 1 − 1
8 2
3
−1/2
(6 − x)
· 2(6 − x) = 1 −
.
8 3 (6 − x) 2 + 4
If T′ (x) = 0, then
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Chapter 4 | Applications of Derivatives
445
6−x
1=
.
8 3 (6 − x) 2 + 4
Therefore,
3 (6 − x) 2 + 4 = 8(6 − x).
(4.6)
Squaring both sides of this equation, we see that if x satisfies this equation, then x must satisfy
9⎡⎣(6 − x) 2 + 4⎤⎦ = 64(6 − x) 2,
which implies
55(6 − x) 2 = 36.
We conclude that if x is a critical point, then x satisfies
(x − 6) 2 = 36 .
55
Therefore, the possibilities for critical points are
x = 6± 6 .
55
Since x = 6 + 6/ 55 is not in the domain, it is not a possibility for a critical point. On the other hand,
x = 6 − 6/ 55 is in the domain. Since we squared both sides of Equation 4.6 to arrive at the possible critical
points, it remains to verify that x = 6 − 6/ 55 satisfies Equation 4.6. Since x = 6 − 6/ 55 does satisfy that
equation, we conclude that x = 6 − 6/ 55 is a critical point, and it is the only one. To justify that the time is
minimized for this value of x, we just need to check the values of T(x) at the endpoints x = 0 and x = 6,
and compare them with the value of T(x) at the critical point x = 6 − 6/ 55. We find that T(0) ≈ 2.108 h
and T(6) ≈ 1.417 h, whereas T ⎛⎝6 − 6/ 55⎞⎠ ≈ 1.368 h. Therefore, we conclude that T has a local minimum at
x ≈ 5.19 mi.
4.33 Suppose the island is 1 mi from shore, and the distance from the cabin to the point on the shore closest
to the island is 15 mi. Suppose a visitor swims at the rate of 2.5 mph and runs at a rate of 6 mph. Let x
denote the distance the visitor will run before swimming, and find a function for the time it takes the visitor to
get from the cabin to the island.
In business, companies are interested in maximizing revenue. In the following example, we consider a scenario in which a
company has collected data on how many cars it is able to lease, depending on the price it charges its customers to rent a
car. Let’s use these data to determine the price the company should charge to maximize the amount of money it brings in.
Example 4.35
Maximizing Revenue
Owners of a car rental company have determined that if they charge customers p dollars per day to rent a
car, where 50 ≤ p ≤ 200, the number of cars n they rent per day can be modeled by the linear function
446
Chapter 4 | Applications of Derivatives
n(p) = 1000 − 5p. If they charge $50 per day or less, they will rent all their cars. If they charge $200 per
day or more, they will not rent any cars. Assuming the owners plan to charge customers between $50 per day and
$200 per day to rent a car, how much should they charge to maximize their revenue?
Solution
Step 1: Let p be the price charged per car per day and let n be the number of cars rented per day. Let R be the
revenue per day.
Step 2: The problem is to maximize R.
Step 3: The revenue (per day) is equal to the number of cars rented per day times the price charged per car per
day—that is, R = n × p.
Step 4: Since the number of cars rented per day is modeled by the linear function n(p) = 1000 − 5p, the
revenue R can be represented by the function
R(p) = n × p = ⎛⎝1000 − 5p⎞⎠ p = −5p 2 + 1000p.
Step 5: Since the owners plan to charge between $50 per car per day and $200 per car per day, the problem is
to find the maximum revenue R(p) for p in the closed interval ⎡⎣50, 200⎤⎦.
Step 6: Since R is a continuous function over the closed, bounded interval ⎡⎣50, 200⎤⎦, it has an absolute
maximum (and an absolute minimum) in that interval. To find the maximum value, look for critical points.
The derivative is R′ (p) = −10p + 1000. Therefore, the critical point is p = 100 When p = 100,
R(100) = $50,000. When p = 50,
R(p) = $37,500. When p = 200,
R(p) = $0. Therefore, the absolute
maximum occurs at p = $100. The car rental company should charge $100 per day per car to maximize
revenue as shown in the following figure.
Figure 4.67 To maximize revenue, a car rental company has to
balance the price of a rental against the number of cars people
will rent at that price.
4.34
A car rental company charges its customers p dollars per day, where 60 ≤ p ≤ 150. It has found that
the number of cars rented per day can be modeled by the linear function n(p) = 750 − 5p. How much should
the company charge each customer to maximize revenue?
Example 4.36
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Chapter 4 | Applications of Derivatives
447
Maximizing the Area of an Inscribed Rectangle
A rectangle is to be inscribed in the ellipse
x 2 + y 2 = 1.
4
What should the dimensions of the rectangle be to maximize its area? What is the maximum area?
Solution
Step 1: For a rectangle to be inscribed in the ellipse, the sides of the rectangle must be parallel to the axes. Let L
be the length of the rectangle and W be its width. Let A be the area of the rectangle.
Figure 4.68 We want to maximize the area of a rectangle inscribed in an
ellipse.
Step 2: The problem is to maximize A.
Step 3: The area of the rectangle is A = LW.
Step 4: Let (x, y) be the corner of the rectangle that lies in the first quadrant, as shown in Figure 4.68. We can
write length L = 2x and width W = 2y. Since
2
x2
and y > 0, we have y = 1 − x . Therefore, the
2
4
4+y =1
area is
2
A = LW = (2x)⎛⎝2y⎞⎠ = 4x 1 − x = 2x 4 − x 2.
4
Step 5: From Figure 4.68, we see that to inscribe a rectangle in the ellipse, the x -coordinate of the corner in
the first quadrant must satisfy 0 < x < 2. Therefore, the problem reduces to looking for the maximum value of
A(x) over the open interval (0, 2). Since A(x) will have an absolute maximum (and absolute minimum) over
the closed interval [0, 2], we consider A(x) = 2x 4 − x 2 over the interval [0, 2]. If the absolute maximum
occurs at an interior point, then we have found an absolute maximum in the open interval.
Step 6: As mentioned earlier, A(x) is a continuous function over the closed, bounded interval [0, 2]. Therefore,
it has an absolute maximum (and absolute minimum). At the endpoints x = 0 and x = 2,
0 < x < 2,
we obtain
A(x) = 0. For
A(x) > 0. Therefore, the maximum must occur at a critical point. Taking the derivative of A(x),
448
Chapter 4 | Applications of Derivatives
1
(−2x)
2 4 − x2
2
= 2 4 − x 2 − 2x
4 − x2
2
= 8 − 4x .
4 − x2
A′(x) = 2 4 − x 2 + 2x ·
To find critical points, we need to find where A′(x) = 0. We can see that if x is a solution of
8 − 4x 2 = 0,
4 − x2
(4.7)
then x must satisfy
8 − 4x 2 = 0.
Therefore, x 2 = 2. Thus, x = ± 2 are the possible solutions of Equation 4.7. Since we are considering x
over the interval [0, 2],
x = 2 is a possibility for a critical point, but x = − 2 is not. Therefore, we check
whether 2 is a solution of Equation 4.7. Since x = 2 is a solution of Equation 4.7, we conclude that 2
is the only critical point of A(x) in the interval [0, 2]. Therefore, A(x) must have an absolute maximum at the
critical point x = 2. To determine the dimensions of the rectangle, we need to find the length L and the width
W. If x = 2 then
y= 1−
( 2) 2
= 1−1 = 1 .
4
2
2
Therefore, the dimensions of the rectangle are L = 2x = 2 2 and W = 2y = 2 = 2. The area of this
2
rectangle is A = LW = (2 2)( 2) = 4.
4.35
Modify the area function A if the rectangle is to be inscribed in the unit circle x 2 + y 2 = 1. What is the
domain of consideration?
Solving Optimization Problems when the Interval Is Not Closed or Is
Unbounded
In the previous examples, we considered functions on closed, bounded domains. Consequently, by the extreme value
theorem, we were guaranteed that the functions had absolute extrema. Let’s now consider functions for which the domain
is neither closed nor bounded.
Many functions still have at least one absolute extrema, even if the domain is not closed or the domain is unbounded. For
example, the function f (x) = x 2 + 4 over (−∞, ∞) has an absolute minimum of 4 at x = 0. Therefore, we can still
consider functions over unbounded domains or open intervals and determine whether they have any absolute extrema. In
the next example, we try to minimize a function over an unbounded domain. We will see that, although the domain of
consideration is (0, ∞), the function has an absolute minimum.
In the following example, we look at constructing a box of least surface area with a prescribed volume. It is not difficult to
show that for a closed-top box, by symmetry, among all boxes with a specified volume, a cube will have the smallest surface
area. Consequently, we consider the modified problem of determining which open-topped box with a specified volume has
the smallest surface area.
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Chapter 4 | Applications of Derivatives
449
Example 4.37
Minimizing Surface Area
A rectangular box with a square base, an open top, and a volume of 216 in.3 is to be constructed. What should
the dimensions of the box be to minimize the surface area of the box? What is the minimum surface area?
Solution
Step 1: Draw a rectangular box and introduce the variable x to represent the length of each side of the square
base; let y represent the height of the box. Let S denote the surface area of the open-top box.
Figure 4.69 We want to minimize the surface area of a
square-based box with a given volume.
Step 2: We need to minimize the surface area. Therefore, we need to minimize S.
Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base.
The area of each of the four vertical sides is x · y. The area of the base is x 2. Therefore, the surface area of the
box is
S = 4xy + x 2.
Step 4: Since the volume of this box is x 2 y and the volume is given as 216 in. 3, the constraint equation is
x 2 y = 216.
Solving the constraint equation for y, we have y = 216
. Therefore, we can write the surface area as a function
2
x
of x only:
⎛ ⎞ 2
S(x) = 4x 216
⎝ x2 ⎠ + x .
2
Therefore, S(x) = 864
x +x .
Step 5: Since we are requiring that x 2 y = 216, we cannot have x = 0. Therefore, we need x > 0. On the
other hand, x is allowed to have any positive value. Note that as x becomes large, the height of the box
y becomes correspondingly small so that x 2 y = 216. Similarly, as x becomes small, the height of the box
becomes correspondingly large. We conclude that the domain is the open, unbounded interval (0, ∞). Note that,
unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute
minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an
absolute minimum over the interval (0, ∞).
Step 6: Note that as x → 0+ ,
S(x) → ∞. Also, as x → ∞,
S(x) → ∞. Since S is a continuous function
450
Chapter 4 | Applications of Derivatives
that approaches infinity at the ends, it must have an absolute minimum at some x ∈ (0, ∞). This minimum must
occur at a critical point of S. The derivative is
S′ (x) = − 864
+ 2x.
x2
Therefore,
S′ (x) = 0 when 2x = 864
. Solving this equation for x,
x2
we obtain
3
3
x 3 = 432,
so
3
x = 432 = 6 2. Since this is the only critical point of S, the absolute minimum must occur at x = 6 2
3
(see Figure 4.70). When x = 6 2,
3
3
y = 216 2 = 3 2 in. Therefore, the dimensions of the box should be
⎛ 3 ⎞
⎝6 2⎠
3
x = 6 2 in. and y = 3 2 in. With these dimensions, the surface area is
3
⎛ 3 ⎞
⎛ 3 ⎞
S⎝6 2⎠ = 864
+ ⎝6 2⎠ = 108 4 in. 2
3
6 2
2
Figure 4.70 We can use a graph to determine the dimensions
of a box of given the volume and the minimum surface area.
4.36
Consider the same open-top box, which is to have volume 216 in. 3. Suppose the cost of the material for
the base is 20 ¢ /in. 2 and the cost of the material for the sides is 30 ¢ /in. 2 and we are trying to minimize the
cost of this box. Write the cost as a function of the side lengths of the base. (Let x be the side length of the base
and y be the height of the box.)
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Chapter 4 | Applications of Derivatives
451
4.7 EXERCISES
For the following exercises,
counterexample, or explanation.
answer
by
proof,
311. When you find the maximum for an optimization
problem, why do you need to check the sign of the
derivative around the critical points?
312. Why do you need to check the endpoints for
optimization problems?
313. True or False. For every continuous nonlinear
function, you can find the value x that maximizes the
function.
320. You have 800 ft of fencing to make a pen for hogs.
If you have a river on one side of your property, what is the
dimension of the rectangular pen that maximizes the area?
321. You need to construct a fence around an area of
1600 ft. What are the dimensions of the rectangular pen to
minimize the amount of material needed?
322. Two poles are connected by a wire that is also
connected to the ground. The first pole is 20 ft tall and
the second pole is 10 ft tall. There is a distance of 30 ft
between the two poles. Where should the wire be anchored
to the ground to minimize the amount of wire needed?
314. True or False. For every continuous nonconstant
function on a closed, finite domain, there exists at least one
x that minimizes or maximizes the function.
For the following exercises, set up and evaluate each
optimization problem.
315.
To carry a suitcase on an airplane, the length
+width + height of the box must be less than or equal
to 62 in. Assuming the height is fixed, show that the
⎛
⎛ ⎞ ⎞
2
maximum volume is V = h⎝31 − ⎝1 ⎠ h⎠ . What height
2
allows you to have the largest volume?
316. You are constructing a cardboard box with the
dimensions 2 m by 4 m. You then cut equal-size squares
from each corner so you may fold the edges. What are the
dimensions of the box with the largest volume?
323. [T] You are moving into a new apartment and notice
there is a corner where the hallway narrows from
8 ft to 6 ft. What is the length of the longest item that can
be carried horizontally around the corner?
317. Find the positive integer that minimizes the sum of
the number and its reciprocal.
318. Find two positive integers such that their sum is 10,
and minimize and maximize the sum of their squares.
For the following exercises, consider the construction of a
pen to enclose an area.
319. You have 400 ft of fencing to construct a
rectangular pen for cattle. What are the dimensions of the
pen that maximize the area?
324.
A
patient’s
70 bpm, 80 bpm, then 120 bpm.
pulse
measures
To determine an
accurate measurement of pulse, the doctor wants to know
what
value
minimizes
the
expression
(x − 70) 2 + (x − 80) 2 + (x − 120) 2 ?
What
value
minimizes it?
452
Chapter 4 | Applications of Derivatives
325. In the previous problem, assume the patient was
nervous during the third measurement, so we only weight
that value half as much as the others. What is the value that
minimizes (x − 70) 2 + (x − 80) 2 + 1 (x − 120) 2 ?
2
326. You can run at a speed of 6 mph and swim at a speed
of 3 mph and are located on the shore, 4 miles east of
an island that is 1 mile north of the shoreline. How far
should you run west to minimize the time needed to reach
the island?
pizzas
for
a
revenue
of
R(x) = ax
and
costs
C(x) = b + cx + dx 2, where x represents the number of
pizzas.
332. Find the profit function for the number of pizzas.
How many pizzas gives the largest profit per pizza?
333.
Assume that R(x) = 10x and C(x) = 2x + x 2.
How many pizzas sold maximizes the profit?
334.
Assume
C(x) = 60 + 3x + 1 x 2.
2
that
How
R(x) = 15x,
and
many
sold
pizzas
maximizes the profit?
For the following problems, consider a lifeguard at a
circular pool with diameter 40 m. He must reach someone
who is drowning on the exact opposite side of the pool, at
position C. The lifeguard swims with a speed v and runs
around the pool at speed w = 3v.
For the following exercises, consider a wire 4 ft long cut
into two pieces. One piece forms a circle with radius r and
the other forms a square of side x.
335. Choose x to maximize the sum of their areas.
336. Choose x to minimize the sum of their areas.
For the following exercises, consider two nonnegative
numbers x and y such that x + y = 10. Maximize and
minimize the quantities.
337. xy
338. x 2 y 2
339. y − 1x
327. Find a function that measures the total amount of
time it takes to reach the drowning person as a function of
the swim angle, θ.
328. Find at what angle θ the lifeguard should swim to
reach the drowning person in the least amount of time.
329. A truck uses gas as g(v) = av + bv , where v
represents the speed of the truck and g represents the
gallons of fuel per mile. At what speed is fuel consumption
minimized?
For the following exercises, consider a limousine that gets
(120 − 2v)
m(v) =
mi/gal at speed v,
5
costs $15/h, and gas is $3.5/gal.
the chauffeur
330. Find the cost per mile at speed v.
331. Find the cheapest driving speed.
For the following exercises, consider a pizzeria that sell
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340. x 2 − y
For the following exercises, draw the given optimization
problem and solve.
341. Find the volume of the largest right circular cylinder
that fits in a sphere of radius 1.
342. Find the volume of the largest right cone that fits in a
sphere of radius 1.
343. Find the area of the largest rectangle that fits into the
y
triangle with sides x = 0, y = 0 and x + = 1.
4
6
344. Find the largest volume of a cylinder that fits into a
cone that has base radius R and height h.
345. Find the dimensions of the closed cylinder volume
V = 16π that has the least amount of surface area.
346. Find the dimensions of a right cone with surface area
S = 4π that has the largest volume.
Chapter 4 | Applications of Derivatives
For the following exercises, consider the points on the
given graphs. Use a calculator to graph the functions.
347. [T] Where is the line y = 5 − 2x closest to the
origin?
348. [T] Where is the line y = 5 − 2x closest to point
(1, 1) ?
349. [T] Where is the parabola y = x 2 closest to point
(2, 0) ?
453
353. You are constructing a box for your cat to sleep in.
The plush material for the square bottom of the box costs
$5/ft 2 and the material for the sides costs $2/ft 2. You
need a box with volume 4 ft 2. Find the dimensions of the
box that minimize cost. Use x to represent the length of the
side of the box.
354. You are building five identical pens adjacent to each
other with a total area of 1000 m 2, as shown in the
following figure. What dimensions should you use to
minimize the amount of fencing?
350. [T] Where is the parabola y = x 2 closest to point
(0, 3) ?
For the following exercises, set up, but do not evaluate,
each optimization problem.
351. A window is composed of a semicircle placed on
top of a rectangle. If you have 20 ft of window-framing
materials for the outer frame, what is the maximum size of
the window you can create? Use r to represent the radius
of the semicircle.
355. You are the manager of an apartment complex with
50 units. When you set rent at $800/month, all
apartments are rented. As you increase rent by
$25/month, one fewer apartment is rented. Maintenance
costs run $50/month for each occupied unit. What is the
rent that maximizes the total amount of profit?
352. You have a garden row of 20 watermelon plants
that produce an average of 30 watermelons apiece. For
any additional watermelon plants planted, the output per
watermelon plant drops by one watermelon. How many
extra watermelon plants should you plant?
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Chapter 4 | Applications of Derivatives
4.8 | L’Hôpital’s Rule
Learning Objectives
4.8.1 Recognize when to apply L’Hôpital’s rule.
4.8.2 Identify indeterminate forms produced by quotients, products, subtractions, and powers,
and apply L’Hôpital’s rule in each case.
4.8.3 Describe the relative growth rates of functions.
In this section, we examine a powerful tool for evaluating limits. This tool, known as L’Hôpital’s rule, uses derivatives to
calculate limits. With this rule, we will be able to evaluate many limits we have not yet been able to determine. Instead of
relying on numerical evidence to conjecture that a limit exists, we will be able to show definitively that a limit exists and to
determine its exact value.
Applying L’Hôpital’s Rule
L’Hôpital’s rule can be used to evaluate limits involving the quotient of two functions. Consider
lim
f (x)
x → a g(x)
.
If xlim
f (x) = L 1 and xlim
g(x) = L 2 ≠ 0, then
→a
→a
lim
f (x)
x → a g(x)
=
L1
.
L2
However, what happens if xlim
f (x) = 0 and xlim
g(x) = 0 ? We call this one of the indeterminate forms, of type 0 .
→a
→a
0
This is considered an indeterminate form because we cannot determine the exact behavior of
f (x)
as x → a without
g(x)
further analysis. We have seen examples of this earlier in the text. For example, consider
2
x.
lim x − 4 and lim sin
x→2 x−2
x→0 x
For the first of these examples, we can evaluate the limit by factoring the numerator and writing
2
(x + 2)(x − 2)
lim x − 4 = lim
= lim (x + 2) = 2 + 2 = 4.
x−2
x→2 x−2
x→2
x→2
x
For lim sin
x we were able to show, using a geometric argument, that
x→0
x
lim sin
x = 1.
x→0
Here we use a different technique for evaluating limits such as these. Not only does this technique provide an easier way to
evaluate these limits, but also, and more important, it provides us with a way to evaluate many other limits that we could
not calculate previously.
The idea behind L’Hôpital’s rule can be explained using local linear approximations. Consider two differentiable functions
f and g such that xlim
f (x) = 0 = xlim
g(x) and such that g′ (a) ≠ 0 For x near a, we can write
→a
→a
f (x) ≈ f (a) + f ′ (a)(x − a)
and
g(x) ≈ g(a) + g′ (a)(x − a).
Therefore,
f (x)
f (a) + f ′ (a)(x − a)
≈
.
g(x) g(a) + g′ (a)(x − a)
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Chapter 4 | Applications of Derivatives
455
Figure 4.71 If lim f (x) = lim g(x), then the ratio f (x)/g(x) is
x→a
x→a
approximately equal to the ratio of their linear approximations near a.
Since f is differentiable at a,
then f is continuous at a,
and therefore f (a) = xlim
f (x) = 0. Similarly,
→a
g(a) = xlim
g(x) = 0. If we also assume that f ′ and g′ are continuous at x = a, then f ′ (a) = xlim
f ′ (x) and
→a
→a
g′ (a) = xlim
g′ (x). Using these ideas, we conclude that
→a
lim
f (x)
x → a g(x)
= xlim
→a
f ′ (x)(x − a)
f ′ (x)
= lim
.
g′ (x)(x − a) x → a g′ (x)
Note that the assumption that f ′ and g′ are continuous at a and g′ (a) ≠ 0 can be loosened. We state L’Hôpital’s rule
formally for the indeterminate form 0 . Also note that the notation 0 does not mean we are actually dividing zero by zero.
0
0
Rather, we are using the notation 0 to represent a quotient of limits, each of which is zero.
0
Theorem 4.12: L’Hôpital’s Rule (0/0 Case)
Suppose f and g are differentiable functions over an open interval containing a,
except possibly at a. If
lim f (x) = 0 and xlim
g(x) = 0, then
→a
x→a
lim
f (x)
x → a g(x)
= xlim
→a
f ′ (x)
,
g′ (x)
assuming the limit on the right exists or is ∞ or −∞. This result also holds if we are considering one-sided limits,
or if a = ∞ and − ∞.
Proof
We provide a proof of this theorem in the special case when f , g, f ′,
and g′ are all continuous over an open
interval containing a. In that case, since xlim
f (x) = 0 = xlim
g(x) and f and g are continuous at a, it follows that
→a
→a
f (a) = 0 = g(a). Therefore,
456
Chapter 4 | Applications of Derivatives
lim
f (x)
x → a g(x)
= xlim
→a
f (x) − f (a)
g(x) − g(a)
since f (a) = 0 = g(a)
f (x) − f (a)
x−a
= xlim
→ a g(x) − g(a)
algebra
=
limit of a quotient
x−a
f (x) − f (a)
lim
x→a x−a
g(x) − g(a)
lim
x→a x−a
f ′ (a)
g′ (a)
lim f ′ (x)
= x→a
lim g′ (x)
x→a
=
= xlim
→a
definition of he derivative
continuity of f ′ and g′
f ′ (x)
.
g′ (x)
limit of a quotient
f
Note that L’Hôpital’s rule states we can calculate the limit of a quotient g by considering the limit of the quotient of the
derivatives
f′
f
. It is important to realize that we are not calculating the derivative of the quotient g .
g′
□
Example 4.38
Applying L’Hôpital’s Rule (0/0 Case)
Evaluate each of the following limits by applying L’Hôpital’s rule.
a.
b.
c.
d.
lim 1 − xcos x
x→0
lim
x→1
sin(πx)
ln x
lim e
x→∞
1/x
−1
1/x
lim sin x2− x
x
x→0
Solution
a. Since the numerator 1 − cos x → 0 and the denominator x → 0, we can apply L’Hôpital’s rule to
evaluate this limit. We have
d (1 − cos x)
dx
d (x)
x→0
dx
lim 1 − xcos x = lim
x→0
= lim sin x
x→0 1
lim (sin x)
= x→0
lim (1)
x→0
= 0 = 0.
1
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Chapter 4 | Applications of Derivatives
457
b. As x → 1, the numerator sin(πx) → 0 and the denominator ln(x) → 0. Therefore, we can apply
L’Hôpital’s rule. We obtain
lim
x→1
sin(πx)
π cos(πx)
= lim
1/x
ln x
x→1
= lim (πx)cos(πx)
x→1
= (π · 1)(−1) = −π.
⎛ ⎞
1/x
c. As x → ∞, the numerator e − 1 → 0 and the denominator ⎝1x ⎠ → 0. Therefore, we can apply
L’Hôpital’s rule. We obtain
1/x
lim e 1− 1 = x lim
x→∞
→∞
⎛ ⎞
e 1/x ⎝−12 ⎠
x
x
⎛−1 ⎞
⎝x2 ⎠
e
= x lim
e 1/x = e 0 = 1 · x lim
→∞
→∞
1/x
− 1.
ln x
d. As x → 0, both the numerator and denominator approach zero. Therefore, we can apply L’Hôpital’s
rule. We obtain
lim sin x2− x = lim cos x − 1 .
2x
x→0
x
x→0
Since the numerator and denominator of this new quotient both approach zero as x → 0, we apply
L’Hôpital’s rule again. In doing so, we see that
lim cos x − 1 = lim −sin x = 0.
2x
x→0 2
x→0
Therefore, we conclude that
lim sin x2− x = 0.
x
x→0
4.37
x .
Evaluate lim tan
x
x→0
We can also use L’Hôpital’s rule to evaluate limits of quotients
f (x)
in which f (x) → ±∞ and g(x) → ±∞. Limits of
g(x)
this form are classified as indeterminate forms of type ∞/∞. Again, note that we are not actually dividing ∞ by ∞.
Since ∞ is not a real number, that is impossible; rather, ∞/∞. is used to represent a quotient of limits, each of which is
∞ or −∞.
Theorem 4.13: L’Hôpital’s Rule (∞/∞ Case)
Suppose f and g are differentiable functions over an open interval containing a, except possibly at a. Suppose
lim f (x) = ∞ (or −∞) and xlim
g(x) = ∞ (or −∞). Then,
→a
x→a
lim
f (x)
x → a g(x)
= xlim
→a
f ′ (x)
,
g′ (x)
assuming the limit on the right exists or is ∞ or −∞. This result also holds if the limit is infinite, if a = ∞ or
458
Chapter 4 | Applications of Derivatives
−∞, or the limit is one-sided.
Example 4.39
Applying L’Hôpital’s Rule (∞/∞ Case)
Evaluate each of the following limits by applying L’Hôpital’s rule.
a.
b.
lim 3x + 5
x → ∞ 2x + 1
ln x
lim cot
x
+
x→0
Solution
3x + 5 and 2x + 1 are first-degree polynomials with positive leading coefficients,
(2x + 1) = ∞. Therefore, we apply L’Hôpital’s rule and obtain
lim (3x + 5) = ∞ and x lim
x→∞
→∞
a. Since
3 + 5/x = lim 3 = 3 .
lim 3x + 5 = x lim
→ ∞ 2x + 1
x → ∞2
2
x → ∞ 2x + 1
Note that this limit can also be calculated without invoking L’Hôpital’s rule. Earlier in the chapter we
showed how to evaluate such a limit by dividing the numerator and denominator by the highest power of
x in the denominator. In doing so, we saw that
3 + 5/x = 3 .
lim 3x + 5 = x lim
→ ∞ 2x + 1
2
x → ∞ 2x + 1
L’Hôpital’s rule provides us with an alternative means of evaluating this type of limit.
b. Here,
lim ln x = −∞ and
x → 0+
lim cot x = ∞. Therefore, we can apply L’Hôpital’s rule and obtain
x → 0+
1/x
ln x = lim
1
lim cot
lim
.
2
x x → 0+ −csc 2 x = x →
0+ −x csc x
x → 0+
Now as x → 0+ ,
csc 2 x → ∞. Therefore, the first term in the denominator is approaching zero and
the second term is getting really large. In such a case, anything can happen with the product. Therefore,
we cannot make any conclusion yet. To evaluate the limit, we use the definition of csc x to write
lim
x → 0+
Now
lim sin 2 x = 0 and
x → 0+
2
x
1
= lim sin
−x .
2
+
−x csc x x → 0
lim x = 0, so we apply L’Hôpital’s rule again. We find
x → 0+
2
x
2sin x cos x = 0 = 0.
lim sin
−x = lim+
+
−1
−1
x→0
x→0
We conclude that
ln x = 0.
lim cot
x
x → 0+
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Chapter 4 | Applications of Derivatives
4.38
459
ln x .
Evaluate x lim
→ ∞ 5x
As mentioned, L’Hôpital’s rule is an extremely useful tool for evaluating limits. It is important to remember, however, that
to apply L’Hôpital’s rule to a quotient
f (x)
f (x)
, it is essential that the limit of
be of the form 0 or ∞/∞. Consider
0
g(x)
g(x)
the following example.
Example 4.40
When L’Hôpital’s Rule Does Not Apply
2
Consider lim x + 5 . Show that the limit cannot be evaluated by applying L’Hôpital’s rule.
x → 1 3x + 4
Solution
Because the limits of the numerator and denominator are not both zero and are not both infinite, we cannot apply
L’Hôpital’s rule. If we try to do so, we get
d ⎛x 2 + 5⎞ = 2x
⎠
dx ⎝
and
d (3x + 4) = 3.
dx
At which point we would conclude erroneously that
2
lim x + 5 = lim 2x = 2 .
3x
+4 x→1 3
3
x→1
However, since lim ⎛⎝x 2 + 5⎞⎠ = 6 and lim (3x + 4) = 7, we actually have
x→1
x→1
2
lim x + 5 = 6 .
7
x → 1 3x + 4
We can conclude that
d ⎛x 2 + 5⎞
2
⎝
⎠
lim x + 5 ≠ lim dx
.
x → 1 3x + 4
x → 1 d (3x + 4)
dx
4.39
Explain why we cannot apply L’Hôpital’s rule to evaluate
x
lim cos
x . Evaluate
x → 0+
x
lim cos
x by other
x → 0+
means.
Other Indeterminate Forms
L’Hôpital’s rule is very useful for evaluating limits involving the indeterminate forms 0 and ∞/∞. However, we can
0
also use L’Hôpital’s rule to help evaluate limits involving other indeterminate forms that arise when evaluating limits. The
expressions 0 · ∞, ∞ − ∞, 1 ∞, ∞ 0, and 0 0 are all considered indeterminate forms. These expressions are not
real numbers. Rather, they represent forms that arise when trying to evaluate certain limits. Next we realize why these are
indeterminate forms and then understand how to use L’Hôpital’s rule in these cases. The key idea is that we must rewrite
460
Chapter 4 | Applications of Derivatives
the indeterminate forms in such a way that we arrive at the indeterminate form 0 or ∞/∞.
0
Indeterminate Form of Type 0 · ∞
⎛
f (x) · g(x)⎞⎠, where f (x) → 0 and g(x) → ∞ (or −∞) as x → a. Since one term
Suppose we want to evaluate xlim
→ a⎝
in the product is approaching zero but the other term is becoming arbitrarily large (in magnitude), anything can happen to
the product. We use the notation 0 · ∞ to denote the form that arises in this situation. The expression 0 · ∞ is considered
indeterminate because we cannot determine without further analysis the exact behavior of the product f (x)g(x) as x → ∞.
For example, let n be a positive integer and consider
f (x) =
As x → ∞,
1
and g(x) = 3x 2.
(x n + 1)
f (x) → 0 and g(x) → ∞. However, the limit as x → ∞ of f (x)g(x) =
3x 2 varies, depending on n.
(x + 1)
n
If n = 2, then x lim
f (x)g(x) = 3. If n = 1, then x lim
f (x)g(x) = ∞. If n = 3, then x lim
f (x)g(x) = 0. Here we
→∞
→∞
→∞
consider another limit involving the indeterminate form 0 · ∞ and show how to rewrite the function as a quotient to use
L’Hôpital’s rule.
Example 4.41
Indeterminate Form of Type 0 · ∞
Evaluate
lim x ln x.
x → 0+
Solution
First, rewrite the function x ln x as a quotient to apply L’Hôpital’s rule. If we write
x ln x = ln x ,
1/x
we see that ln x → −∞ as x → 0+ and 1x → ∞ as x → 0+ . Therefore, we can apply L’Hôpital’s rule and
obtain
lim ln x = lim
x → 0+ 1/x
x → 0+
d (ln x)
dx
d (1/x)
dx
2
= lim 1/x = lim (−x) = 0.
x → 0+ −1/x
x → 0+
We conclude that
lim x ln x = 0.
x → 0+
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Chapter 4 | Applications of Derivatives
461
Figure 4.72 Finding the limit at x = 0 of the function
f (x) = x ln x.
4.40
Evaluate lim x cot x.
x→0
Indeterminate Form of Type ∞ − ∞
Another type of indeterminate form is ∞ − ∞. Consider the following example. Let n be a positive integer and let
f (x) = 3x n and g(x) = 3x 2 + 5. As x → ∞,
f (x) → ∞ and g(x) → ∞. We are interested in
lim ⎛ f (x) − g(x)⎞⎠.
x → ∞⎝
Depending on whether f (x) grows faster, g(x) grows faster, or they grow at the same rate, as we see next, anything can
happen in this limit. Since f (x) → ∞ and g(x) → ∞, we write ∞ − ∞ to denote the form of this limit. As with our
other indeterminate forms, ∞ − ∞ has no meaning on its own and we must do more analysis to determine the value of the
limit. For example, suppose the exponent n in the function f (x) = 3x n is n = 3, then
⎛ 3
3x − 3x 2 − 5⎞⎠ = ∞.
lim ⎛ f (x) − g(x)⎞⎠ = x lim
→ ∞⎝
x → ∞⎝
On the other hand, if n = 2, then
⎛ 2
lim ⎛ f (x) − g(x)⎞⎠ = x lim
3x − 3x 2 − 5⎞⎠ = −5.
→ ∞⎝
x → ∞⎝
However, if n = 1, then
⎛
lim ⎛ f (x) − g(x)⎞⎠ = x lim
3x − 3x 2 − 5⎞⎠ = −∞.
→ ∞⎝
x → ∞⎝
Therefore, the limit cannot be determined by considering only ∞ − ∞. Next we see how to rewrite an expression involving
the indeterminate form ∞ − ∞ as a fraction to apply L’Hôpital’s rule.
Example 4.42
Indeterminate Form of Type ∞ − ∞
462
Chapter 4 | Applications of Derivatives
Evaluate
⎛1
2
x → 0+ ⎝ x
lim
1 ⎞.
− tan
x
⎠
Solution
By combining the fractions, we can write the function as a quotient. Since the least common denominator is
x 2 tan x, we have
2
1 − 1 = (tan x) − x .
x 2 tan x
x 2 tan x
As x → 0+ ,
the numerator tan x − x 2 → 0 and the denominator x 2 tan x → 0. Therefore, we can apply
L’Hôpital’s rule. Taking the derivatives of the numerator and the denominator, we have
⎛
2 ⎞
⎝sec x⎠ − 2x
(tan x) − x 2
=
lim
.
2
2
2
x → 0+ x tan x
x → 0+ x sec x + 2x tan x
lim
As x → 0+ ,
⎛
2 ⎞
⎝sec x⎠ −
2x → 1 and x 2 sec 2 x + 2x tan x → 0. Since the denominator is positive as x
approaches zero from the right, we conclude that
⎛
2 ⎞
⎝sec x⎠ −
lim
2x
2
2
x → 0+ x sec x + 2x tan x
Therefore,
⎛1
2
x → 0+ ⎝ x
lim
4.41
Evaluate
= ∞.
1 ⎞ = ∞.
− tan
x
⎠
⎞
⎛
lim ⎝1x − 1 ⎠.
sin x
x → 0+
Another type of indeterminate form that arises when evaluating limits involves exponents. The expressions 0 0,
∞ 0, and
1 ∞ are all indeterminate forms. On their own, these expressions are meaningless because we cannot actually evaluate these
expressions as we would evaluate an expression involving real numbers. Rather, these expressions represent forms that arise
when finding limits. Now we examine how L’Hôpital’s rule can be used to evaluate limits involving these indeterminate
forms.
Since L’Hôpital’s rule applies to quotients, we use the natural logarithm function and its properties to reduce a problem
evaluating a limit involving exponents to a related problem involving a limit of a quotient. For example, suppose we want
g(x)
to evaluate xlim
f (x)
and we arrive at the indeterminate form ∞ 0. (The indeterminate forms 0 0 and 1 ∞ can be
→a
handled similarly.) We proceed as follows. Let
y = f (x)
Then,
⎛
ln y = ln⎝ f (x)
g(x)
.
g(x)⎞
⎛
⎞
⎠ = g(x)ln⎝ f (x)⎠.
Therefore,
⎡
⎛
⎞⎤
lim ⎡⎣ln(y)⎤⎦ = xlim
⎣g(x)ln⎝ f (x)⎠⎦.
→a
x→a
Since xlim
f (x) = ∞,
→a
we know that xlim
ln⎛ f (x)⎞⎠ = ∞. Therefore, xlim
g(x)ln⎛⎝ f (x)⎞⎠ is of the indeterminate form
→a ⎝
→a
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Chapter 4 | Applications of Derivatives
463
0 · ∞, and we can use the techniques discussed earlier to rewrite the expression g(x)ln⎛⎝ f (x)⎞⎠ in a form so that we can
apply L’Hôpital’s rule. Suppose xlim
g(x)ln⎛⎝ f (x)⎞⎠ = L, where L may be ∞ or −∞. Then
→a
lim ⎡⎣ln(y)⎤⎦ = L.
x→a
Since the natural logarithm function is continuous, we conclude that
y⎞ = L,
ln⎛⎝xlim
→a ⎠
which gives us
lim y = xlim
f (x)
→a
x→a
g(x)
= e L.
Example 4.43
Indeterminate Form of Type ∞0
Evaluate x lim
x 1/x.
→∞
Solution
Let y = x 1/x. Then,
ln⎛⎝x 1/x⎞⎠ = 1x ln x = lnxx .
ln x . Applying L’Hôpital’s rule, we obtain
We need to evaluate x lim
→∞ x
ln x = lim 1/x = 0.
lim ln y = x lim
→∞ x
x→∞ 1
x→∞
Therefore, x lim
ln y = 0. Since the natural logarithm function is continuous, we conclude that
→∞
y⎞ = 0,
ln⎛⎝x lim
→∞ ⎠
which leads to
ln x = e 0 = 1.
lim y = x lim
→∞ x
x→∞
Hence,
lim x 1/x = 1.
x→∞
4.42
Evaluate x lim
x
→∞
1/ln(x)
.
Example 4.44
Indeterminate Form of Type 0 0
464
Chapter 4 | Applications of Derivatives
Evaluate
lim x sin x.
x → 0+
Solution
Let
y = x sin x.
Therefore,
ln y = ln⎛⎝x sin x⎞⎠ = sin x ln x.
lim sin x ln x. Since
We now evaluate
x → 0+
lim sin x = 0 and
x → 0+
lim ln x = −∞, we have the indeterminate
x → 0+
form 0 · ∞. To apply L’Hôpital’s rule, we need to rewrite sin x ln x as a fraction. We could write
sin x ln x = sin x
1/ln x
or
ln x .
sin x ln x = ln x = csc
x
1/sin x
Let’s consider the first option. In this case, applying L’Hôpital’s rule, we would obtain
cos x
lim sin x ln x = lim sin x = lim
= lim ⎛⎝−x(ln x) 2 cos x⎞⎠.
⎛
2⎞
+
+
+
1/ln
x
x→0
x→0
x → 0 −1/⎝x(ln x) ⎠
x → 0+
Unfortunately, we not only have another expression involving the indeterminate form 0 · ∞, but the new limit
is even more complicated to evaluate than the one with which we started. Instead, we try the second option. By
writing
ln x ,
sin x ln x = ln x = csc
x
1/sin x
and applying L’Hôpital’s rule, we obtain
1/x
ln x = lim
lim sin x ln x = lim csc
lim x csc−1
x
−csc x cot x = x →
x cot x .
x → 0+
x → 0+
0+
x → 0+
Using the fact that csc x =
1 and cot x = cos x , we can rewrite the expression on the right-hand side as
sin x
sin x
2
⎞
⎡ sin x
⎤ ⎛
x
sin x ⎞ ⎛
lim −sin
x
cos
x = lim+ ⎣ x · (−tan x)⎦ = ⎝ lim+ x ⎠ · ⎝ lim+ (−tan x)⎠ = 1 · 0 = 0.
+
x→0
We conclude that
x→0
x→0
x→0
⎛
⎞
lim y = 0 and we have
+
⎝x →
0 ⎠
lim ln y = 0. Therefore, ln
x → 0+
lim y = lim x sin x = e 0 = 1.
x → 0+
x → 0+
Hence,
lim x sin x = 1.
x → 0+
4.43
Evaluate
lim x x.
x → 0+
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Chapter 4 | Applications of Derivatives
465
Growth Rates of Functions
Suppose the functions f and g both approach infinity as x → ∞. Although the values of both functions become
arbitrarily large as the values of x become sufficiently large, sometimes one function is growing more quickly than the
other. For example, f (x) = x 2 and g(x) = x 3 both approach infinity as x → ∞. However, as shown in the following
table, the values of x 3 are growing much faster than the values of x 2.
x
10
100
1000
10,000
f(x) = x2
100
10,000
1,000,000
100,000,000
g(x) = x3
1000
1,000,000
1,000,000,000
1,000,000,000,000
Table 4.21 Comparing the Growth Rates of x 2 and x 3
In fact,
3
x 2 = lim 1 = 0.
lim x = x lim
x = ∞. or, equivalently, x lim
→∞
→∞ 3
x → ∞x
x
x → ∞ x2
As a result, we say x 3 is growing more rapidly than x 2 as x → ∞. On the other hand, for f (x) = x 2 and
g(x) = 3x 2 + 4x + 1, although the values of g(x) are always greater than the values of f (x) for x > 0, each value of
g(x) is roughly three times the corresponding value of f (x) as x → ∞, as shown in the following table. In fact,
x2
lim
x → ∞ 3x 2 + 4x + 1
= 1.
3
x
10
100
1000
10,000
f(x) = x2
100
10,000
1,000,000
100,000,000
g(x) = 3x2 + 4x + 1
341
30,401
3,004,001
300,040,001
Table 4.22 Comparing the Growth Rates of x 2 and 3x 2 + 4x + 1
In this case, we say that x 2 and 3x 2 + 4x + 1 are growing at the same rate as x → ∞.
More generally, suppose f and g are two functions that approach infinity as x → ∞. We say g grows more rapidly than
f as x → ∞ if
lim
g(x)
x → ∞ f (x)
= ∞; or, equivalently, x lim
→∞
On the other hand, if there exists a constant M ≠ 0 such that
lim
f (x)
x → ∞ g(x)
we say f and g grow at the same rate as x → ∞.
= M,
f (x)
= 0.
g(x)
466
Chapter 4 | Applications of Derivatives
Next we see how to use L’Hôpital’s rule to compare the growth rates of power, exponential, and logarithmic functions.
Example 4.45
Comparing the Growth Rates of ln(x), x2, and e x
⎛ f (x) ⎞
⎝ g(x) ⎠.
For each of the following pairs of functions, use L’Hôpital’s rule to evaluate x lim
→∞
a.
f (x) = x 2 and g(x) = e x
b.
f (x) = ln(x) and g(x) = x 2
Solution
⎡ 2⎤
x . We obtain
a. Since x lim
x 2 = ∞ and x lim
e x, we can use L’Hôpital’s rule to evaluate x lim
→∞
→∞
→ ∞⎣ e x ⎦
2
2x .
lim x = x lim
→ ∞ex
x → ∞ex
Since x lim
2x = ∞ and x lim
e x = ∞, we can apply L’Hôpital’s rule again. Since
→∞
→∞
2 = 0,
lim 2x = x lim
→ ∞e x
x → ∞ex
we conclude that
2
lim x = 0.
x → ∞ex
Therefore, e x grows more rapidly than x 2 as x → ∞ (See Figure 4.73 and Table 4.23).
Figure 4.73 An exponential function grows at a faster rate
than a power function.
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Chapter 4 | Applications of Derivatives
467
x
5
10
15
20
x2
25
100
225
400
ex
148
22,026
3,269,017
485,165,195
Table 4.23
Growth rates of a power function and an exponential function.
ln x . We
ln x = ∞ and x lim
x 2 = ∞, we can use L’Hôpital’s rule to evaluate x lim
b. Since x lim
→∞
→∞
→ ∞ x2
obtain
1/x = lim 1 = 0.
lim ln2x = x lim
→ ∞ 2x
x → ∞ 2x 2
x
x→∞
Thus, x 2 grows more rapidly than ln x as x → ∞ (see Figure 4.74 and Table 4.24).
Figure 4.74 A power function grows at a faster rate than a
logarithmic function.
x
10
100
1000
10,000
ln(x)
2.303
4.605
6.908
9.210
x2
100
10,000
1,000,000
100,000,000
Table 4.24
Growth rates of a power function and a logarithmic function
468
Chapter 4 | Applications of Derivatives
4.44
Compare the growth rates of x 100 and 2 x.
Using the same ideas as in Example 4.45a. it is not difficult to show that e x grows more rapidly than x p for any p > 0.
In Figure 4.75 and Table 4.25, we compare e x with x 3 and x 4 as x → ∞.
Figure 4.75 The exponential function e x grows faster than x p for any p > 0. (a) A comparison of e x with
x 3. (b) A comparison of e x with x 4.
x
5
10
15
20
x3
125
1000
3375
8000
x4
625
10,000
50,625
160,000
ex
148
22,026
3,269,017
485,165,195
Table 4.25 An exponential function grows at a faster rate than
any power function
Similarly, it is not difficult to show that x p grows more rapidly than ln x for any p > 0. In Figure 4.76 and Table 4.26,
we compare ln x with 3 x and
x.
Figure 4.76 The function y = ln(x) grows more slowly than
x p for any p > 0 as x → ∞.
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Chapter 4 | Applications of Derivatives
469
x
10
100
1000
10,000
ln(x)
2.303
4.605
6.908
9.210
x
2.154
4.642
10
21.544
x
3.162
10
31.623
100
3
Table 4.26 A logarithmic function grows at a slower rate
than any root function
470
Chapter 4 | Applications of Derivatives
4.8 EXERCISES
For the following exercises, evaluate the limit.
372.
x
e .
356. Evaluate the limit x lim
→∞ x
373.
(1 + x) n − 1
x
x→0
374.
(1 + x) n − 1 − nx
x→0
x2
x
e .
357. Evaluate the limit x lim
→∞ k
x
ln x .
358. Evaluate the limit x lim
→∞ k
x
x−a .
359. Evaluate the limit xlim
→ a x2 − a2
x−a .
360. Evaluate the limit xlim
→a 3
x − a3
x−a .
361. Evaluate the limit xlim
n
→a n
375.
376.
377.
x −a
For the following exercises, determine whether you can
apply L’Hôpital’s rule directly. Explain why or why not.
Then, indicate if there is some way you can alter the limit
so you can apply L’Hôpital’s rule.
362.
lim x 2 ln x
x → 0+
363. x lim
x 1/x
→∞
364.
365.
lim x 2/x
x→0
2
lim x
x → 0 1/x
x
378.
379.
380.
381.
382.
For the following exercises, evaluate the limits with either
L’Hôpital’s rule or previously learned methods.
385.
lim sin x −3 tan x
x
x→0
1−x
lim 1 + x −
x
x→0
x
lim e − x2 − 1
x→0
x
lim tan x
x
x→0
lim x − 1
ln x
x→1
lim (x + 1) 1/x
x→0
3
lim x − x
x
−
1
x→1
lim x 2x
x → 0+
lim sin x2− x
x
x→0
lim x ln⎛⎝x 4⎞⎠
x → 0+
2
lim x − 9
x−3
(x − e x)
386. x lim
→∞
2
lim x − 9
x+3
387. x lim
x 2 e −x
→∞
x→3
x→3
(1 + x) −2 − 1
369. lim
x
x→0
388.
lim cos x
x → π/2 π
2−x
389.
370.
lim
⎛ ⎞
384.
368.
lim
383. x lim
x sin⎝1x ⎠
→∞
e
366. x lim
→∞ x
367.
lim x − 1
sin x
x→1
x−π
371. xlim
→ π sin x
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390.
x
2x
lim 3 −
x
x→0
lim 1 + 1/x
x → 0 1 − 1/x
lim (1 − tan x)cot x
x → π/4
Chapter 4 | Applications of Derivatives
391. x lim
xe 1/x
→∞
392.
393.
394.
lim x 1/cos x
x→0
lim x 1/x
x→0
lim ⎛⎝1 − 1x ⎞⎠
x
x→0
⎛
1 − 1x ⎞⎠
395. x lim
→ ∞⎝
x
For the following exercises, use a calculator to graph the
function and estimate the value of the limit, then use
L’Hôpital’s rule to find the limit directly.
x
396. [T] lim e x− 1
x→0
⎛ ⎞
397. [T] lim x sin⎝1x ⎠
x→0
398. [T] lim
x−1
x → 1 1 − cos(πx)
399. [T] lim e
x→1
(x − 1)
−1
x−1
(x − 1) 2
x → 1 ln x
400. [T] lim
1 + cos x
401. [T] xlim
→ π sin x
⎛
⎞
402. [T] lim ⎝csc x − 1x ⎠
x→0
403. [T]
404. [T]
lim tan(x x)
x → 0+
lim ln x
sin x
x → 0+
−x
x
405. [T] lim e −x e
x→0
471
472
Chapter 4 | Applications of Derivatives
4.9 | Newton’s Method
Learning Objectives
4.9.1
4.9.2
4.9.3
4.9.4
Describe the steps of Newton’s method.
Explain what an iterative process means.
Recognize when Newton’s method does not work.
Apply iterative processes to various situations.
In many areas of pure and applied mathematics, we are interested in finding solutions to an equation of the form f (x) = 0.
For most functions, however, it is difficult—if not impossible—to calculate their zeroes explicitly. In this section, we take
a look at a technique that provides a very efficient way of approximating the zeroes of functions. This technique makes use
of tangent line approximations and is behind the method used often by calculators and computers to find zeroes.
Describing Newton’s Method
Consider the task of finding the solutions of f (x) = 0. If f is the first-degree polynomial f (x) = ax + b, then the
2
solution of f (x) = 0 is given by the formula x = − b
a . If f is the second-degree polynomial f (x) = ax + bx + c,
the solutions of f (x) = 0 can be found by using the quadratic formula. However, for polynomials of degree 3 or more,
finding roots of f becomes more complicated. Although formulas exist for third- and fourth-degree polynomials, they are
quite complicated. Also, if f is a polynomial of degree 5 or greater, it is known that no such formulas exist. For example,
consider the function
f (x) = x 5 + 8x 4 + 4x 3 − 2x − 7.
No formula exists that allows us to find the solutions of f (x) = 0. Similar difficulties exist for nonpolynomial functions.
For example, consider the task of finding solutions of tan(x) − x = 0. No simple formula exists for the solutions of this
equation. In cases such as these, we can use Newton’s method to approximate the roots.
Newton’s method makes use of the following idea to approximate the solutions of f (x) = 0. By sketching a graph of
f , we can estimate a root of f (x) = 0. Let’s call this estimate x 0. We then draw the tangent line to f at x 0. If
f ′ (x 0) ≠ 0, this tangent line intersects the x -axis at some point ⎛⎝x 1, 0⎞⎠. Now let x 1 be the next approximation to the
actual root. Typically, x 1 is closer than x 0 to an actual root. Next we draw the tangent line to f at x 1. If f ′ (x 1) ≠ 0,
this tangent line also intersects the x -axis, producing another approximation, x 2. We continue in this way, deriving a list
of approximations: x 0, x 1, x 2 ,…. Typically, the numbers x 0, x 1, x 2 ,… quickly approach an actual root x * , as shown
in the following figure.
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Chapter 4 | Applications of Derivatives
473
Figure 4.77 The approximations x 0, x 1, x 2 ,… approach the actual root x * . The
approximations are derived by looking at tangent lines to the graph of f .
Now let’s look at how to calculate the approximations x 0, x 1, x 2 ,…. If x 0 is our first approximation, the approximation
x 1 is defined by letting ⎛⎝x 1, 0⎞⎠ be the x -intercept of the tangent line to f at x 0. The equation of this tangent line is given
by
y = f (x 0) + f ′ (x 0)(x − x 0).
Therefore, x 1 must satisfy
f (x 0) + f ′ (x 0)(x 1 − x 0) = 0.
Solving this equation for x 1, we conclude that
x1 = x0 −
f (x 0)
.
f ′(x 0)
Similarly, the point ⎛⎝x 2, 0⎞⎠ is the x -intercept of the tangent line to f at x 1. Therefore, x 2 satisfies the equation
x2 = x1 −
f (x 1)
.
f ′(x 1)
In general, for n > 0, x n satisfies
xn = xn − 1 −
f (x n − 1)
.
f ′(x n − 1)
Next we see how to make use of this technique to approximate the root of the polynomial f (x) = x 3 − 3x + 1.
(4.8)
474
Chapter 4 | Applications of Derivatives
Example 4.46
Finding a Root of a Polynomial
Use Newton’s method to approximate a root of f (x) = x 3 − 3x + 1 in the interval [1, 2]. Let x 0 = 2 and find
x 1, x 2, x 3, x 4, and x 5.
Solution
From Figure 4.78, we see that f has one root over the interval (1, 2). Therefore x 0 = 2 seems like
a reasonable first approximation. To find the next approximation, we use Equation 4.8. Since
f (x) = x 3 − 3x + 1, the derivative is f ′ (x) = 3x 2 − 3. Using Equation 4.8 with n = 1 (and a calculator
that displays 10 digits), we obtain
x1 = x0 −
f (x 0)
f (2)
=2−
= 2 − 3 ≈ 1.666666667.
9
f ′(2)
f ′(x 0)
To find the next approximation, x 2, we use Equation 4.8 with n = 2 and the value of x 1 stored on the
calculator. We find that
x2 = x1 =
f (x 1)
≈ 1.548611111.
f ′(x 1)
Continuing in this way, we obtain the following results:
x 1 ≈ 1.666666667
x 2 ≈ 1.548611111
x 3 ≈ 1.532390162
x 4 ≈ 1.532088989
x 5 ≈ 1.532088886
x 6 ≈ 1.532088886.
We note that we obtained the same value for x 5 and x 6. Therefore, any subsequent application of Newton’s
method will most likely give the same value for x n.
Figure 4.78 The function f (x) = x 3 − 3x + 1 has one root
over the interval [1, 2].
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Chapter 4 | Applications of Derivatives
4.45
475
Letting x 0 = 0, let’s use Newton’s method to approximate the root of f (x) = x 3 − 3x + 1 over the
interval [0, 1] by calculating x 1 and x 2.
Newton’s method can also be used to approximate square roots. Here we show how to approximate
be modified to approximate the square root of any positive number.
2. This method can
Example 4.47
Finding a Square Root
Use Newton’s method to approximate
2 (Figure 4.79). Let f (x) = x 2 − 2, let x 0 = 2, and calculate
x 1, x 2, x 3, x 4, x 5. (We note that since f (x) = x 2 − 2 has a zero at
reasonable choice to approximate
2, the initial value x 0 = 2 is a
2.)
Solution
For f (x) = x 2 − 2, f ′ (x) = 2x. From Equation 4.8, we know that
xn = xn − 1 −
f (x n − 1)
f ′(x n − 1)
x2 n − 1 − 2
= xn − 1 −
2x n − 1
= 1 xn − 1 + x 1
2
n−1
⎛
⎞
= 1 ⎝x n − 1 + x 2 ⎠.
2
n−1
Therefore,
⎛
x 1 = 1 ⎝x 0 +
2
⎛
x 2 = 1 ⎝x 1 +
2
2 ⎞ 1⎛
2⎞
x 0 ⎠ = 2 ⎝2 + 2 ⎠ = 1.5
2 ⎞
2 ⎞ 1⎛
x 1 ⎠ = 2 ⎝1.5 + 1.5 ⎠ ≈ 1.416666667.
Continuing in this way, we find that
x 1 = 1.5
x 2 ≈ 1.416666667
x 3 ≈ 1.414215686
x 4 ≈ 1.414213562
x 5 ≈ 1.414213562.
Since we obtained the same value for x 4 and x 5, it is unlikely that the value x n will change on any subsequent
application of Newton’s method. We conclude that 2 ≈ 1.414213562.
476
Chapter 4 | Applications of Derivatives
Figure 4.79 We can use Newton’s method to find
4.46
2.
Use Newton’s method to approximate 3 by letting f (x) = x 2 − 3 and x 0 = 3. Find x 1 and x 2.
When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation
⎡ f (x) ⎤
, we can rewrite Equation 4.8 as
f ′ (x) ⎦
by using the same formula. In particular, by defining the function F(x) = x − ⎣
x n = F(x n − 1). This type of process, where each x n is defined in terms of x n − 1 by repeating the same function, is an
example of an iterative process. Shortly, we examine other iterative processes. First, let’s look at the reasons why Newton’s
method could fail to find a root.
Failures of Newton’s Method
Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’s
method might fail include the following:
1. At one of the approximations x n, the derivative f ′ is zero at x n, but f (x n) ≠ 0. As a result, the tangent line of
f at x n does not intersect the x -axis. Therefore, we cannot continue the iterative process.
2. The approximations x 0, x 1, x 2 ,… may approach a different root. If the function f has more than one root, it is
possible that our approximations do not approach the one for which we are looking, but approach a different root
(see Figure 4.80). This event most often occurs when we do not choose the approximation x 0 close enough to the
desired root.
3. The approximations may fail to approach a root entirely. In Example 4.48, we provide an example of a function
and an initial guess x 0 such that the successive approximations never approach a root because the successive
approximations continue to alternate back and forth between two values.
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Chapter 4 | Applications of Derivatives
477
Figure 4.80 If the initial guess x 0 is too far from the root sought, it may lead
to approximations that approach a different root.
Example 4.48
When Newton’s Method Fails
Consider the function f (x) = x 3 − 2x + 2. Let x 0 = 0. Show that the sequence x 1, x 2 ,… fails to approach a
root of f .
Solution
For f (x) = x 3 − 2x + 2, the derivative is f ′ (x) = 3x 2 − 2. Therefore,
x1 = x0 −
f (x 0)
f (0)
=0−
= − 2 = 1.
−2
f ′ (x 0)
f ′ (0)
x2 = x1 −
f (x 1)
f (1)
=1−
= 1 − 1 = 0.
1
f ′(x 1)
f ′ (1)
In the next step,
Consequently, the numbers x 0, x 1, x 2 ,… continue to bounce back and forth between 0 and 1 and never get
closer to the root of f which is over the interval [−2, −1] (see Figure 4.81). Fortunately, if we choose an
initial approximation x 0 closer to the actual root, we can avoid this situation.
478
Chapter 4 | Applications of Derivatives
Figure 4.81 The approximations continue to alternate
between 0 and 1 and never approach the root of f .
4.47
For f (x) = x 3 − 2x + 2, let x 0 = −1.5 and find x 1 and x 2.
From Example 4.48, we see that Newton’s method does not always work. However, when it does work, the sequence of
approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a
root found using Newton’s method are included in texts on numerical analysis.
Other Iterative Processes
As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type of
iterative process.
Consider a function F and an initial number x 0. Define the subsequent numbers x n by the formula x n = F(x n − 1). This
process is an iterative process that creates a list of numbers x 0, x 1, x 2 ,…, x n ,…. This list of numbers may approach a
finite number x * as n gets larger, or it may not. In Example 4.49, we see an example of a function F and an initial
guess x 0 such that the resulting list of numbers approaches a finite value.
Example 4.49
Finding a Limit for an Iterative Process
Let F(x) = 1 x + 4 and let x 0 = 0. For all n ≥ 1, let x n = F(x n − 1). Find the values x 1, x 2, x 3, x 4, x 5.
2
Make a conjecture about what happens to this list of numbers x 1, x 2, x 3 …, x n ,… as n → ∞. If the list of
numbers x 1, x 2, x 3 ,… approaches a finite number x * , then x * satisfies x * = F(x * ), and x * is called
a fixed point of F.
Solution
If x 0 = 0, then
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Chapter 4 | Applications of Derivatives
479
x 1 = 1 (0) + 4 = 4
2
1
x 2 = (4) + 4 = 6
2
x 3 = 1 (6) + 4 = 7
2
x 4 = 1 (7) + 4 = 7.5
2
x 5 = 1 (7.5) + 4 = 7.75
2
x 6 = 1 (7.75) + 4 = 7.875
2
x 7 = 1 (7.875) + 4 = 7.9375
2
x 8 = 1 (7.9375) + 4 = 7.96875
2
x 9 = 1 (7.96875) + 4 = 7.984375.
2
From this list, we conjecture that the values x n approach 8.
Figure 4.82 provides a graphical argument that the values approach 8 as n → ∞. Starting at the point
(x 0, x 0), we draw a vertical line to the point ⎛⎝x 0, F(x 0)⎞⎠. The next number in our list is x 1 = F(x 0). We use
x 1 to calculate x 2. Therefore, we draw a horizontal line connecting (x 0, x 1) to the point (x 1, x 1) on the line
y = x, and then draw a vertical line connecting (x 1, x 1) to the point ⎛⎝x 1, F(x 1)⎞⎠. The output F(x 1) becomes
x 2. Continuing in this way, we could create an infinite number of line segments. These line segments are trapped
between the lines F(x) = x + 4 and y = x. The line segments get closer to the intersection point of these two
2
lines, which occurs when x = F(x). Solving the equation x = x + 4, we conclude they intersect at x = 8.
2
Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers x 0, x 1, x 2 ,…
approaches x * = 8 as n → ∞.
Figure 4.82 This iterative process approaches the value
x * = 8.
480
Chapter 4 | Applications of Derivatives
Consider the function F(x) = 1 x + 6. Let x 0 = 0 and let x n = F(x n − 1) for n ≥ 2. Find
3
x 1, x 2, x 3, x 4, x 5. Make a conjecture about what happens to the list of numbers x 1, x 2, x 3 ,…x n ,… as
n → ∞.
4.48
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Chapter 4 | Applications of Derivatives
481
Iterative Processes and Chaos
Iterative processes can yield some very interesting behavior. In this section, we have seen several examples of iterative
processes that converge to a fixed point. We also saw in Example 4.48 that the iterative process bounced back and
forth between two values. We call this kind of behavior a 2 -cycle. Iterative processes can converge to cycles with
various periodicities, such as 2 − cycles, 4 − cycles (where the iterative process repeats a sequence of four values),
8-cycles, and so on.
Some iterative processes yield what mathematicians call chaos. In this case, the iterative process jumps from value to
value in a seemingly random fashion and never converges or settles into a cycle. Although a complete exploration of
chaos is beyond the scope of this text, in this project we look at one of the key properties of a chaotic iterative process:
sensitive dependence on initial conditions. This property refers to the concept that small changes in initial conditions
can generate drastically different behavior in the iterative process.
Probably the best-known example of chaos is the Mandelbrot set (see Figure 4.83), named after Benoit Mandelbrot
(1924–2010), who investigated its properties and helped popularize the field of chaos theory. The Mandelbrot set is
usually generated by computer and shows fascinating details on enlargement, including self-replication of the set.
Several colorized versions of the set have been shown in museums and can be found online and in popular books on
the subject.
Figure 4.83 The Mandelbrot set is a well-known example of a set of points generated by the
iterative chaotic behavior of a relatively simple function.
In this project we use the logistic map
f (x) = rx(1 − x), where x ∈ [0, 1] and r > 0
as the function in our iterative process. The logistic map is a deceptively simple function; but, depending on the value
of r, the resulting iterative process displays some very interesting behavior. It can lead to fixed points, cycles, and
even chaos.
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Chapter 4 | Applications of Derivatives
To visualize the long-term behavior of the iterative process associated with the logistic map, we will use a tool called a
cobweb diagram. As we did with the iterative process we examined earlier in this section, we first draw a vertical line
from the point ⎛⎝x 0, 0⎞⎠ to the point ⎛⎝x 0, f (x 0)⎞⎠ = (x 0, x 1). We then draw a horizontal line from that point to the point
(x 1, x 1), then draw a vertical line to ⎛⎝x 1, f (x 1)⎞⎠ = (x 1, x 2), and continue the process until the long-term behavior
of the system becomes apparent. Figure 4.84 shows the long-term behavior of the logistic map when r = 3.55 and
x 0 = 0.2. (The first 100 iterations are not plotted.) The long-term behavior of this iterative process is an 8 -cycle.
Figure 4.84 A cobweb diagram for f (x) = 3.55x(1 − x) is
presented here. The sequence of values results in an 8 -cycle.
1. Let r = 0.5 and choose x 0 = 0.2. Either by hand or by using a computer, calculate the first 10 values in the
sequence. Does the sequence appear to converge? If so, to what value? Does it result in a cycle? If so, what
kind of cycle (for example, 2 − cycle, 4 − cycle.) ?
2. What happens when r = 2 ?
3. For r = 3.2 and r = 3.5, calculate the first 100 sequence values. Generate a cobweb diagram for each
iterative process. (Several free applets are available online that generate cobweb diagrams for the logistic map.)
What is the long-term behavior in each of these cases?
4. Now let r = 4. Calculate the first 100 sequence values and generate a cobweb diagram. What is the longterm behavior in this case?
5. Repeat the process for r = 4, but let x 0 = 0.201. How does this behavior compare with the behavior for
x 0 = 0.2 ?
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Chapter 4 | Applications of Derivatives
483
4.9 EXERCISES
For the following exercises, write Newton’s formula as
x n + 1 = F(x n) for solving f (x) = 0.
using Newton’s method and a computer or calculator.
Choose any initial guess x 0 that is not the exact root.
406.
f (x) = x 2 + 1
422. x 2 − 10 = 0
407.
f (x) = x 3 + 2x + 1
423. x 4 − 100 = 0
408.
f (x) = sin x
424. x 2 − x = 0
409.
f (x) = e x
425. x 3 − x = 0
410.
f (x) = x 3 + 3xe x
426. x + 5cos(x) = 0
For the following exercises, solve f (x) = 0 using the
iteration x n + 1 = x n − c f (x n),
which differs slightly
from Newton’s method. Find a c that works and a c that
fails to converge, with the exception of c = 0.
411.
412.
2
f (x) = x − 4, with x 0 = 0
2
f (x) = x − 4x + 3, with x 0 = 2
413. What is the value of “c” for Newton’s method?
For the following exercises, start at
a. x 0 = 0.6 and
⎞
⎛
427. x + tan(x) = 0, choose x 0 ∈ ⎝− π , π ⎠
2 2
428.
1 =2
1−x
429. 1 + x + x 2 + x 3 + x 4 = 2
430. x 3 + (x + 1) 3 = 10 3
431. x = sin 2 (x)
For the following exercises, use Newton’s method to find
the fixed points of the function where f (x) = x; round to
three decimals.
b. x 0 = 2.
432. sin x
Compute x 1 and x 2 using the specified iterative method.
⎞
⎛
433. tan(x) on x = ⎝π , 3π ⎠
2 2
414. x n + 1 = x n 2 − 1
2
434. e x − 2
415. x n + 1 = 2x n ⎛⎝1 − x n⎞⎠
435. ln(x) + 2
416. x n + 1 = x n
Newton’s method can be used to find maxima and minima
of functions in addition to the roots. In this case apply
Newton’s method to the derivative function f ′ (x) to find
417. x n + 1 = 1
xn
418. x n + 1 = 3x n ⎛⎝1 − x n⎞⎠
2
419. x n + 1 = x n + x n − 2
420. x n + 1 = 1 x n − 1
2
421. x n + 1 = |x n|
For the following exercises, solve to four decimal places
its roots, instead of the original function. For the following
exercises, consider the formulation of the method.
436. To find candidates for maxima and minima, we need
to find the critical points f ′ (x) = 0. Show that to solve for
the critical points of a function f (x), Newton’s method is
f ′ (x )
n .
given by x n + 1 = x n −
f ″(x n)
437. What additional restrictions are necessary on the
function f ?
484
Chapter 4 | Applications of Derivatives
For the following exercises, use Newton’s method to find
the location of the local minima and/or maxima of the
following functions; round to three decimals.
438. Minimum of f (x) = x 2 + 2x + 4
439. Minimum of f (x) = 3x 3 + 2x 2 − 16
440. Minimum of f (x) = x 2 e x
454. Why would you use the secant method over
Newton’s method? What are the necessary restrictions on
f?
For the following exercises, use both Newton’s method
and the secant method to calculate a root for the following
equations. Use a calculator or computer to calculate how
many iterations of each are needed to reach within three
decimal places of the exact answer. For the secant method,
use the first guess from Newton’s method.
441. Maximum of f (x) = x + 1x
455.
f (x) = x 2 + 2x + 1, x 0 = 1
442. Maximum of f (x) = x 3 + 10x 2 + 15x − 2
456.
f (x) = x 2, x 0 = 1
3
x
443. Maximum of f (x) = x −
x
457.
f (x) = sin x, x 0 = 1
458.
f (x) = e x − 1, x 0 = 2
459.
f (x) = x 3 + 2x + 4, x 0 = 0
444.
Minimum of
2
f (x) = x sin x,
closest non-zero
minimum to x = 0
445. Minimum of f (x) = x 4 + x 3 + 3x 2 + 12x + 6
For the following exercises, use the specified method to
solve the equation. If it does not work, explain why it does
not work.
In the following exercises, consider Kepler’s equation
regarding planetary orbits, M = E − εsin(E), where M
is the mean anomaly, E is eccentric anomaly, and ε
measures eccentricity.
446. Newton’s method, x 2 + 2 = 0
460. Use Newton’s method to solve for the eccentric
anomaly E when the mean anomaly M = π and the
447. Newton’s method, 0 = e x
eccentricity of the orbit
448. Newton’s method, 0 = 1 + x 2 starting at x 0 = 0
449. Solving x n + 1 = −x n 3 starting at x 0 = −1
For the following exercises, use the secant method, an
alternative iterative method to Newton’s method. The
formula is given by
x n = x n − 1 − f (x n − 1)
xn − 1 − xn − 2
.
f (x n − 1) − f (x n − 2)
450. Find a root to 0 = x 2 − x − 3 accurate to three
decimal places.
451. Find a root to 0 = sin x + 3x accurate to four
decimal places.
x
452. Find a root to 0 = e − 2 accurate to four decimal
places.
453.
Find a root to ln(x + 2) = 1 accurate to four
2
decimal places.
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3
ε = 0.25; round to three
decimals.
461.
Use Newton’s method to solve for the eccentric
anomaly E when the mean anomaly M = 3π and the
2
eccentricity of the orbit ε = 0.8; round to three decimals.
The following two exercises consider a bank investment.
The initial investment is $10,000. After 25 years, the
investment has tripled to $30,000.
462. Use Newton’s method to determine the interest rate
if the interest was compounded annually.
463. Use Newton’s method to determine the interest rate
if the interest was compounded continuously.
464. The cost for printing a book can be given by the
⎛ ⎞
equation C(x) = 1000 + 12x + ⎝1 ⎠x 2/3. Use Newton’s
2
method to find the break-even point if the printer sells each
book for $20.
Chapter 4 | Applications of Derivatives
485
4.10 | Antiderivatives
Learning Objectives
4.10.1
4.10.2
4.10.3
4.10.4
Find the general antiderivative of a given function.
Explain the terms and notation used for an indefinite integral.
State the power rule for integrals.
Use antidifferentiation to solve simple initial-value problems.
At this point, we have seen how to calculate derivatives of many functions and have been introduced to a variety of their
applications. We now ask a question that turns this process around: Given a function f , how do we find a function with
the derivative f and why would we be interested in such a function?
We answer the first part of this question by defining antiderivatives. The antiderivative of a function f is a function with a
derivative f . Why are we interested in antiderivatives? The need for antiderivatives arises in many situations, and we look
at various examples throughout the remainder of the text. Here we examine one specific example that involves rectilinear
motion. In our examination in Derivatives of rectilinear motion, we showed that given a position function s(t) of an
object, then its velocity function v(t) is the derivative of s(t) —that is, v(t) = s′ (t). Furthermore, the acceleration a(t)
is the derivative of the velocity v(t) —that is, a(t) = v′ (t) = s″(t). Now suppose we are given an acceleration function
a, but not the velocity function v or the position function s. Since a(t) = v′ (t), determining the velocity function
requires us to find an antiderivative of the acceleration function. Then, since v(t) = s′ (t), determining the position
function requires us to find an antiderivative of the velocity function. Rectilinear motion is just one case in which the
need for antiderivatives arises. We will see many more examples throughout the remainder of the text. For now, let’s look
at the terminology and notation for antiderivatives, and determine the antiderivatives for several types of functions. We
examine various techniques for finding antiderivatives of more complicated functions later in the text (Introduction to
Techniques of Integration (http://cnx.org/content/m53654/latest/) ).
The Reverse of Differentiation
At this point, we know how to find derivatives of various functions. We now ask the opposite question. Given a function
f , how can we find a function with derivative f ? If we can find a function F derivative f , we call F an antiderivative
of f .
Definition
A function F is an antiderivative of the function f if
F′ (x) = f (x)
for all x in the domain of f .
Consider the function f (x) = 2x. Knowing the power rule of differentiation, we conclude that F(x) = x 2 is an
antiderivative of f since F′ (x) = 2x. Are there any other antiderivatives of f ? Yes; since the derivative of any constant
C is zero, x 2 + C is also an antiderivative of 2x. Therefore, x 2 + 5 and x 2 − 2 are also antiderivatives. Are there any
others that are not of the form x 2 + C for some constant C ? The answer is no. From Corollary 2 of the Mean Value
Theorem, we know that if F and G are differentiable functions such that F′ (x) = G′ (x), then F(x) − G(x) = C for
some constant C. This fact leads to the following important theorem.
486
Chapter 4 | Applications of Derivatives
Theorem 4.14: General Form of an Antiderivative
Let F be an antiderivative of f over an interval I. Then,
i. for each constant C, the function F(x) + C is also an antiderivative of f over I;
ii. if G is an antiderivative of f over I, there is a constant C for which G(x) = F(x) + C over I.
In other words, the most general form of the antiderivative of f over I is F(x) + C.
We use this fact and our knowledge of derivatives to find all the antiderivatives for several functions.
Example 4.50
Finding Antiderivatives
For each of the following functions, find all antiderivatives.
a.
f (x) = 3x 2
b.
f (x) = 1x
c.
f (x) = cos x
d.
f (x) = e x
Solution
a. Because
d ⎛x 3⎞ = 3x 2
dx ⎝ ⎠
then F(x) = x 3 is an antiderivative of 3x 2. Therefore, every antiderivative of 3x 2 is of the form
x 3 + C for some constant C, and every function of the form x 3 + C is an antiderivative of 3x 2.
b. Let f (x) = ln|x|. For x > 0, f (x) = ln(x) and
d (ln x) = 1 .
x
dx
For x < 0, f (x) = ln(−x) and
d ⎛⎝ln(−x)⎞⎠ = − 1 = 1 .
−x x
dx
Therefore,
d (ln|x|) = 1 .
x
dx
Thus, F(x) = ln|x| is an antiderivative of 1x . Therefore, every antiderivative of 1x is of the form
ln|x| + C for some constant C and every function of the form ln|x| + C is an antiderivative of 1x .
c. We have
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Chapter 4 | Applications of Derivatives
487
d (sin x) = cos x,
dx
so F(x) = sin x is an antiderivative of cos x. Therefore, every antiderivative of cos x is of the form
sin x + C for some constant C and every function of the form sin x + C is an antiderivative of cos x.
d. Since
d (e x) = e x,
dx
then F(x) = e x is an antiderivative of e x. Therefore, every antiderivative of e x is of the form e x + C
for some constant C and every function of the form e x + C is an antiderivative of e x.
Find all antiderivatives of f (x) = sin x.
4.49
Indefinite Integrals
We now look at the formal notation used to represent antiderivatives and examine some of their properties. These properties
allow us to find antiderivatives of more complicated functions. Given a function f , we use the notation f ′ (x) or
df
dx
to denote the derivative of f . Here we introduce notation for antiderivatives. If F is an antiderivative of f , we say that
F(x) + C is the most general antiderivative of f and write
∫ f (x)dx = F(x) + C.
The symbol
∫
is called an integral sign, and
∫ f (x)dx
is called the indefinite integral of f .
Definition
Given a function f , the indefinite integral of f , denoted
∫ f (x)dx,
is the most general antiderivative of f . If F is an antiderivative of f , then
∫ f (x)dx = F(x) + C.
The expression f (x) is called the integrand and the variable x is the variable of integration.
Given the terminology introduced in this definition, the act of finding the antiderivatives of a function f is usually referred
to as integrating f .
For a function f and an antiderivative F, the functions F(x) + C, where C is any real number, is often referred to as
the family of antiderivatives of f . For example, since x 2 is an antiderivative of 2x and any antiderivative of 2x is of the
form x 2 + C, we write
488
Chapter 4 | Applications of Derivatives
∫ 2x dx = x 2 + C.
The collection of all functions of the form x 2 + C, where C is any real number, is known as the family of antiderivatives
of 2x. Figure 4.85 shows a graph of this family of antiderivatives.
Figure 4.85 The family of antiderivatives of 2x consists of all functions of the
form x 2 + C, where C is any real number.
For some functions, evaluating indefinite integrals follows directly from properties of derivatives. For example, for
n ≠ −1,
n+1
∫ x n dx = xn + 1 + C,
which comes directly from
d ⎛x n + 1 ⎞ = (n + 1) x n = x n.
n+1
dx ⎝ n + 1 ⎠
This fact is known as the power rule for integrals.
Theorem 4.15: Power Rule for Integrals
For n ≠ −1,
n+1
∫ x n dx = xn + 1 + C.
Evaluating indefinite integrals for some other functions is also a straightforward calculation. The following table lists the
indefinite integrals for several common functions. A more complete list appears in Appendix B.
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Chapter 4 | Applications of Derivatives
489
Differentiation Formula
Indefinite Integral
d (k) = 0
dx
∫ kdx = ∫ kx 0 dx = kx + C
d (x n) = nx n − 1
dx
∫ x n dn = xn + 1 + C
d (ln|x|) = 1
x
dx
∫ 1x dx = ln|x| + C
d (e x) = e x
dx
∫ e x dx = e x + C
d (sin x) = cos x
dx
∫ cos x dx = sin x + C
d (cos x) = −sin x
dx
∫ sin x dx = −cos x + C
d (tan x) = sec 2 x
dx
∫ sec 2 x dx = tan x + C
d (csc x) = −csc x cot x
dx
∫ csc x cot x dx = −csc x + C
d (sec x) = sec x tan x
dx
∫ sec x tan x dx = sec x + C
d (cot x) = −csc 2 x
dx
∫ csc 2 x dx = −cot x + C
d ⎛sin −1 x⎞ =
1
⎠
dx ⎝
1 − x2
∫
1
= sin −1 x + C
2
1−x
d ⎛tan −1 x⎞ = 1
⎠
dx ⎝
1 + x2
∫
1 dx = tan −1 x + C
1 + x2
d ⎛sec −1 |x|⎞ =
1
⎠
dx ⎝
2
x x −1
∫
n+1
1
2
x x −1
Table 4.27 Integration Formulas
From the definition of indefinite integral of f , we know
∫ f (x)dx = F(x) + C
for n ≠ −1
dx = sec −1 |x| + C
490
Chapter 4 | Applications of Derivatives
if and only if F is an antiderivative of f . Therefore, when claiming that
∫ f (x)dx = F(x) + C
it is important to check whether this statement is correct by verifying that F′ (x) = f (x).
Example 4.51
Verifying an Indefinite Integral
Each of the following statements is of the form
∫ f (x)dx = F(x) + C.
Verify that each statement is correct by
showing that F′ (x) = f (x).
2
a.
∫ (x + e x)dx = x2
b.
∫ xe x dx = xe x − e x + C
+ ex + C
Solution
a. Since
d ⎛ x 2 + e x + C ⎞ = x + e x,
⎠
dx ⎝ 2
the statement
2
∫ (x + e x)dx = x2
+ ex + C
is correct.
2
Note that we are verifying an indefinite integral for a sum. Furthermore, x and e x are antiderivatives
2
x
of x and e , respectively, and the sum of the antiderivatives is an antiderivative of the sum. We discuss
this fact again later in this section.
b. Using the product rule, we see that
d (xe x − e x + C) = e x + xe x − e x = xe x.
dx
Therefore, the statement
∫ xe x dx = xe x − e x + C
is correct.
Note that we are verifying an indefinite integral for a product. The antiderivative xe x − e x is not
a product of the antiderivatives. Furthermore, the product of antiderivatives, x 2 e x /2 is not an
antiderivative of xe x since
d ⎛x 2 e x ⎞ = xe x + x 2 e x ≠ xe x.
2
dx ⎝ 2 ⎠
In general, the product of antiderivatives is not an antiderivative of a product.
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Chapter 4 | Applications of Derivatives
4.50
Verify that
491
∫ x cos x dx = x sin x + cos x + C.
In Table 4.27, we listed the indefinite integrals for many elementary functions. Let’s now turn our attention to evaluating
indefinite integrals for more complicated functions. For example, consider finding an antiderivative of a sum f + g.
In Example 4.51a. we showed that an antiderivative of the sum x + e x is given by the sum
⎛x 2 ⎞ x
⎝ 2 ⎠ + e —that is, an
antiderivative of a sum is given by a sum of antiderivatives. This result was not specific to this example. In general, if F
and G are antiderivatives of any functions f and g, respectively, then
d (F(x) + G(x)) = F′ (x) + G′ (x) = f (x) + g(x).
dx
Therefore, F(x) + G(x) is an antiderivative of f (x) + g(x) and we have
∫
⎛
⎝
f (x) + g(x)⎞⎠dx = F(x) + G(x) + C.
∫
⎛
⎝
f (x) − g(x)⎞⎠dx = F(x) − G(x) + C.
Similarly,
In addition, consider the task of finding an antiderivative of k f (x), where k is any real number. Since
d ⎛⎝k f (x)⎞⎠ = k d F(x) = kF′ (x)
dx
dx
for any real number k, we conclude that
∫ k f (x)dx = kF(x) + C.
These properties are summarized next.
Theorem 4.16: Properties of Indefinite Integrals
Let F and G be antiderivatives of f and g, respectively, and let k be any real number.
Sums and Differences
∫
⎛
⎝
f (x)±g(x)⎞⎠dx = F(x)±G(x) + C
Constant Multiples
∫ k f (x)dx = kF(x) + C
From this theorem, we can evaluate any integral involving a sum, difference, or constant multiple of functions with
antiderivatives that are known. Evaluating integrals involving products, quotients, or compositions is more complicated (see
Example 4.51b. for an example involving an antiderivative of a product.) We look at and address integrals involving these
more complicated functions in Introduction to Integration. In the next example, we examine how to use this theorem to
calculate the indefinite integrals of several functions.
Example 4.52
Evaluating Indefinite Integrals
492
Chapter 4 | Applications of Derivatives
Evaluate each of the following indefinite integrals:
a.
∫ ⎛⎝5x 3 − 7x 2 + 3x + 4⎞⎠dx
b.
∫x
c.
∫
d.
∫ tan x cos x dx
2
+ 43 x dx
x
4 dx
1 + x2
Solution
a. Using Properties of Indefinite Integrals, we can integrate each of the four terms in the integrand
separately. We obtain
∫ ⎛⎝5x 3 − 7x 2 + 3x + 4⎞⎠dx = ∫ 5x 3 dx − ∫ 7x 2 dx + ∫ 3x dx + ∫ 4dx.
From the second part of Properties of Indefinite Integrals, each coefficient can be written in front of
the integral sign, which gives
∫ 5x 3 dx − ∫ 7x 2 dx + ∫ 3x dx + ∫ 4dx = 5∫ x 3 dx − 7∫ x 2 dx + 3∫ x dx + 4∫ 1dx.
Using the power rule for integrals, we conclude that
∫ ⎛⎝5x 3 − 7x 2 + 3x + 4⎞⎠dx = 54 x 4 − 73 x 3 + 32 x 2 + 4x + C.
b. Rewrite the integrand as
x 2 + 43 x = x 2 + 43 x = 0.
x
x
x
Then, to evaluate the integral, integrate each of these terms separately. Using the power rule, we have
∫ ⎛⎝x +
x
4 ⎞dx = ∫ x dx + 4∫ x −2/3dx
2/3
⎠
(−2/3) + 1
= 1 x 2 + 4 ⎛ ⎞1 x
+C
−2 + 1
2
⎝ 3 ⎠
= 1 x 2 + 12x 1/3 + C.
2
c. Using Properties of Indefinite Integrals, write the integral as
4∫
1 dx.
1 + x2
Then, use the fact that tan −1 (x) is an antiderivative of
∫
⎛
⎝1
1
to conclude that
+ x 2⎞⎠
4 dx = 4tan −1 (x) + C.
1 + x2
d. Rewrite the integrand as
sin x cos x = sin x.
tan x cos x = cos
x
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Chapter 4 | Applications of Derivatives
493
Therefore,
∫ tan x cos x = ∫ sin x = −cos x + C.
4.51
Evaluate
∫ ⎛⎝4x 3 − 5x 2 + x − 7⎞⎠dx.
Initial-Value Problems
We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in
the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential
equations.
A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation
dy
= f (x)
dx
(4.9)
is a simple example of a differential equation. Solving this equation means finding a function y with a derivative f .
Therefore, the solutions of Equation 4.9 are the antiderivatives of f . If F is one antiderivative of f , every function of
the form y = F(x) + C is a solution of that differential equation. For example, the solutions of
dy
= 6x 2
dx
are given by
y = ∫ 6x 2 dx = 2x 3 + C.
Sometimes we are interested in determining whether a particular solution curve passes through a certain point (x 0, y 0)
—that is, y(x 0) = y 0. The problem of finding a function y that satisfies a differential equation
dy
= f (x)
dx
(4.10)
y(x 0) = y 0
(4.11)
with the additional condition
is an example of an initial-value problem. The condition y(x 0) = y 0 is known as an initial condition. For example,
looking for a function y that satisfies the differential equation
dy
= 6x 2
dx
and the initial condition
y(1) = 5
is an example of an initial-value problem. Since the solutions of the differential equation are y = 2x 3 + C, to find a
function y that also satisfies the initial condition, we need to find C such that y(1) = 2(1) 3 + C = 5. From this equation,
we see that C = 3, and we conclude that y = 2x 3 + 3 is the solution of this initial-value problem as shown in the
following graph.
494
Chapter 4 | Applications of Derivatives
Figure 4.86 Some of the solution curves of the differential equation
dy
= 6x 2
dx
are displayed. The function y = 2x 3 + 3 satisfies the differential equation and the
initial condition y(1) = 5.
Example 4.53
Solving an Initial-Value Problem
Solve the initial-value problem
dy
= sin x, y(0) = 5.
dx
Solution
First we need to solve the differential equation. If
dy
= sin x, then
dx
y = ∫ sin(x)dx = −cos x + C.
Next we need to look for a solution y that satisfies the initial condition. The initial condition y(0) = 5 means
we need a constant C such that −cos x + C = 5. Therefore,
C = 5 + cos(0) = 6.
The solution of the initial-value problem is y = −cos x + 6.
4.52
Solve the initial value problem
dy
= 3x −2, y(1) = 2.
dx
Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car.
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Chapter 4 | Applications of Derivatives
495
We are interested in how long it takes for the car to stop. Recall that the velocity function v(t) is the derivative of a position
function s(t), and the acceleration a(t) is the derivative of the velocity function. In earlier examples in the text, we could
calculate the velocity from the position and then compute the acceleration from the velocity. In the next example we work
the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function
to determine the position function.
Example 4.54
Decelerating Car
A car is traveling at the rate of 88 ft/sec (60 mph) when the brakes are applied. The car begins decelerating at a
constant rate of 15 ft/sec2.
a. How many seconds elapse before the car stops?
b. How far does the car travel during that time?
Solution
a. First we introduce variables for this problem. Let t be the time (in seconds) after the brakes are first
applied. Let a(t) be the acceleration of the car (in feet per seconds squared) at time t. Let v(t) be the
velocity of the car (in feet per second) at time t. Let s(t) be the car’s position (in feet) beyond the point
where the brakes are applied at time t.
The car is traveling at a rate of 88 ft/sec. Therefore, the initial velocity is v(0) = 88 ft/sec. Since the car
is decelerating, the acceleration is
a(t) = −15 ft/s 2.
The acceleration is the derivative of the velocity,
v′ (t) = 15.
Therefore, we have an initial-value problem to solve:
v′ (t) = −15, v(0) = 88.
Integrating, we find that
v(t) = −15t + C.
Since v(0) = 88, C = 88. Thus, the velocity function is
v(t) = −15t + 88.
To find how long it takes for the car to stop, we need to find the time t such that the velocity is zero.
Solving −15t + 88 = 0, we obtain t = 88 sec.
15
b. To find how far the car travels during this time, we need to find the position of the car after 88 sec. We
15
know the velocity v(t) is the derivative of the position s(t). Consider the initial position to be s(0) = 0.
Therefore, we need to solve the initial-value problem
s′ (t) = −15t + 88, s(0) = 0.
496
Chapter 4 | Applications of Derivatives
Integrating, we have
s(t) = − 15 t 2 + 88t + C.
2
Since s(0) = 0, the constant is C = 0. Therefore, the position function is
s(t) = − 15 t 2 + 88t.
2
⎛ ⎞
After t = 88 sec, the position is s⎝88 ⎠ ≈ 258.133 ft.
15
15
4.53 Suppose the car is traveling at the rate of 44 ft/sec. How long does it take for the car to stop? How far
will the car travel?
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Chapter 4 | Applications of Derivatives
497
4.10 EXERCISES
F(x)
483.
f (x) = sin 2 (x)cos(x)
484.
f (x) = 0
F(x) = 5x 3 + 2x 2 + 3x + 1, f (x) = 15x 2 + 4x + 3
485.
f (x) = 1 csc 2 (x) + 12
2
x
466. F(x) = x 2 + 4x + 1, f (x) = 2x + 4
486.
f (x) = csc x cot x + 3x
467. F(x) = x 2 e x, f (x) = e x ⎛⎝x 2 + 2x⎞⎠
487.
f (x) = 4csc x cot x − sec x tan x
468. F(x) = cos x, f (x) = −sin x
488.
f (x) = 8sec x(sec x − 4tan x)
469. F(x) = e x, f (x) = e x
489.
f (x) = 1 e −4x + sin x
2
For the following exercises, find the antiderivative of the
function.
For the following exercises, evaluate the integral.
For the following exercises, show that
are
antiderivatives of f (x).
465.
470.
f (x) = 12 + x
x
490.
∫ (−1)dx
471.
f (x) = e x − 3x 2 + sin x
491.
∫ sin x dx
472.
f (x) = e x + 3x − x 2
492.
∫ (4x +
473.
f (x) = x − 1 + 4sin(2x)
493.
∫ 3x
494.
∫ (sec x tan x + 4x)dx
495.
∫ ⎛⎝4
496.
∫ ⎛⎝x −1/3 − x 2/3⎞⎠dx
497.
∫ 14x
∫ (e x + e −x)dx
For the following exercises, find the antiderivative F(x) of
each function f (x).
474.
f (x) = 5x 4 + 4x 5
475.
f (x) = x + 12x 2
476.
f (x) = 1
x
3
x)dx
2
+ 2 dx
x2
x + 4 x⎞⎠dx
3
+ 2x + 1 dx
x3
477.
f (x) = ( x)
478.
f (x) = x 1/3 + (2x) 1/3
498.
479.
1/3
f (x) = x 2/3
x
For the following exercises, solve the initial value problem.
480.
481.
482.
499.
f ′ (x) = x −3, f (1) = 1
500.
f ′ (x) = x + x 2, f (0) = 2
501.
⎛ ⎞
f ′ (x) = cos x + sec 2 (x), f ⎝π ⎠ = 2 + 2
2
4
502.
f ′ (x) = x 3 − 8x 2 + 16x + 1, f (0) = 0
f (x) = 2sin(x) + sin(2x)
f (x) = sec 2 (x) + 1
f (x) = sin x cos x
498
503.
Chapter 4 | Applications of Derivatives
2
f ′ (x) = 22 − x , f (1) = 0
2
x
For the following exercises, find two possible functions f
given the second- or third-order derivatives.
2
For the following exercises, determine whether the
statement is true or false. Either prove it is true or find a
counterexample if it is false.
521. If f (x) is the antiderivative of v(x), then 2 f (x) is
the antiderivative of 2v(x).
504.
f ″(x) = x + 2
505.
f ″(x) = e −x
the antiderivative of v(2x).
506.
f ″(x) = 1 + x
523. If f (x) is the antiderivative of v(x), then f (x) + 1
507.
f ‴(x) = cos x
508.
f ‴(x) = 8e
−2x
522. If f (x) is the antiderivative of v(x), then f (2x) is
is the antiderivative of v(x) + 1.
− sin x
509. A car is being driven at a rate of 40 mph when the
brakes are applied. The car decelerates at a constant rate of
10 ft/sec2. How long before the car stops?
510. In the preceding problem, calculate how far the car
travels in the time it takes to stop.
511. You are merging onto the freeway, accelerating at a
constant rate of 12 ft/sec2. How long does it take you to
reach merging speed at 60 mph?
512. Based on the previous problem, how far does the car
travel to reach merging speed?
513. A car company wants to ensure its newest model can
stop in 8 sec when traveling at 75 mph. If we assume
constant deceleration, find the value of deceleration that
accomplishes this.
514. A car company wants to ensure its newest model can
stop in less than 450 ft when traveling at 60 mph. If we
assume constant deceleration, find the value of deceleration
that accomplishes this.
For the following exercises, find the antiderivative of the
function, assuming F(0) = 0.
515. [T] f (x) = x 2 + 2
516. [T] f (x) = 4x − x
517. [T] f (x) = sin x + 2x
518. [T] f (x) = e x
519. [T] f (x) =
1
(x + 1) 2
520. [T] f (x) = e −2x + 3x 2
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524. If f (x) is the antiderivative of v(x), then ⎛⎝ f (x)⎞⎠ 2
is the antiderivative of (v(x)) 2.
Chapter 4 | Applications of Derivatives
499
CHAPTER 4 REVIEW
KEY TERMS
absolute extremum if f has an absolute maximum or absolute minimum at c, we say f has an absolute extremum
at c
absolute maximum if f (c) ≥ f (x) for all x in the domain of f , we say f has an absolute maximum at c
absolute minimum if f (c) ≤ f (x) for all x in the domain of f , we say f has an absolute minimum at c
antiderivative a function F such that F′ (x) = f (x) for all x in the domain of f is an antiderivative of f
concave down if f is differentiable over an interval I and f ′ is decreasing over I, then f is concave down over
I
concave up if f is differentiable over an interval I and f ′ is increasing over I, then f is concave up over I
concavity the upward or downward curve of the graph of a function
concavity test suppose f is twice differentiable over an interval I; if f ″ > 0 over I, then f is concave up over I;
if f ″ < 0 over I, then f is concave down over I
critical point if f ′(c) = 0 or f ′(c) is undefined, we say that c is a critical point of f
differential the differential dx is an independent variable that can be assigned any nonzero real number; the differential
dy is defined to be dy = f ′(x)dx
differential form given a differentiable function y = f ′(x), the equation dy = f ′(x)dx is the differential form of the
derivative of y with respect to x
end behavior the behavior of a function as x → ∞ and x → −∞
extreme value theorem if f is a continuous function over a finite, closed interval, then f has an absolute maximum
and an absolute minimum
Fermat’s theorem if f has a local extremum at c, then c is a critical point of f
first derivative test let f be a continuous function over an interval I containing a critical point c such that f is
differentiable over I except possibly at c; if f ′ changes sign from positive to negative as x increases through c,
then f has a local maximum at c; if f ′ changes sign from negative to positive as x increases through c, then f
has a local minimum at c; if f ′ does not change sign as x increases through c, then f does not have a local
extremum at c
f (x) = L or x →
lim
f (x) = L, then y = L is a horizontal asymptote of f
horizontal asymptote if x lim
→∞
−∞
indefinite integral the most general antiderivative of f (x) is the indefinite integral of f ; we use the notation
∫ f (x)dx
to denote the indefinite integral of f
indeterminate forms when evaluating a limit, the forms 0 ,
0
∞/∞,
0 · ∞,
∞ − ∞,
0 0,
∞ 0, and 1 ∞ are
considered indeterminate because further analysis is required to determine whether the limit exists and, if so, what its
value is
infinite limit at infinity a function that becomes arbitrarily large as x becomes large
inflection point if f is continuous at c and f changes concavity at c, the point ⎛⎝c, f (c)⎞⎠ is an inflection point of f
500
Chapter 4 | Applications of Derivatives
initial value problem
a problem that requires finding a function y that satisfies the differential equation
dy
= f (x)
dx
together with the initial condition y(x 0) = y 0
iterative process process in which a list of numbers x 0, x 1, x 2, x 3 … is generated by starting with a number x 0 and
defining x n = F(x n − 1) for n ≥ 1
limit at infinity the limiting value, if it exists, of a function as x → ∞ or x → −∞
linear approximation the linear function L(x) = f (a) + f ′(a)(x − a) is the linear approximation of f at x = a
local extremum if f has a local maximum or local minimum at c, we say f has a local extremum at c
local maximum if there exists an interval I such that f (c) ≥ f (x) for all x ∈ I, we say f has a local maximum at
c
local minimum if there exists an interval I such that f (c) ≤ f (x) for all x ∈ I, we say f has a local minimum at c
L’Hôpital’s rule if
f and g are differentiable functions over an interval a,
lim f (x) = 0 = xlim
g(x) or xlim
f (x) and xlim
g(x) are infinite, then xlim
→a
→a
→a
→a
x→a
except possibly at a,
and
f (x)
f ′ (x)
= lim
, assuming the
g(x) x → a g′ (x)
limit on the right exists or is ∞ or −∞
mean value theorem if f is continuous over [a, b] and differentiable over (a, b), then there exists c ∈ (a, b) such
that
f ′ (c) =
f (b) − f (a)
b−a
Newton’s method method for approximating roots of f (x) = 0; using an initial guess x 0; each subsequent
f (x
)
n−1
approximation is defined by the equation x n = x n − 1 −
f ′(x n − 1)
oblique asymptote the line y = mx + b if f (x) approaches it as x → ∞ or x → −∞
optimization problems problems that are solved by finding the maximum or minimum value of a function
percentage error the relative error expressed as a percentage
propagated error the error that results in a calculated quantity f (x) resulting from a measurement error dx
related rates are rates of change associated with two or more related quantities that are changing over time
relative error
Δq
given an absolute error Δq for a particular quantity, q is the relative error.
rolle’s theorem if f is continuous over [a, b] and differentiable over (a, b), and if f (a) = f (b), then there exists
c ∈ (a, b) such that f ′ (c) = 0
second derivative test suppose f ′ (c) = 0 and f ″ is continuous over an interval containing c; if f ″(c) > 0, then
f has a local minimum at c; if f ″(c) < 0, then f has a local maximum at c; if f ″(c) = 0, then the test is
inconclusive
tangent line approximation (linearization) since the linear approximation of f at x = a is defined using the
equation of the tangent line, the linear approximation of f at x = a is also known as the tangent line approximation
to f at x = a
KEY EQUATIONS
• Linear approximation
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Chapter 4 | Applications of Derivatives
501
L(x) = f (a) + f ′(a)(x − a)
• A differential
dy = f ′(x)dx.
KEY CONCEPTS
4.1 Related Rates
• To solve a related rates problem, first draw a picture that illustrates the relationship between the two or more related
quantities that are changing with respect to time.
• In terms of the quantities, state the information given and the rate to be found.
• Find an equation relating the quantities.
• Use differentiation, applying the chain rule as necessary, to find an equation that relates the rates.
• Be sure not to substitute a variable quantity for one of the variables until after finding an equation relating the rates.
4.2 Linear Approximations and Differentials
• A differentiable function y = f (x) can be approximated at a by the linear function
L(x) = f (a) + f ′(a)(x − a).
• For a function y = f (x), if x changes from a to a + dx, then
dy = f ′(x)dx
is an approximation for the change in y. The actual change in y is
Δy = f (a + dx) − f (a).
• A measurement error dx can lead to an error in a calculated quantity f (x). The error in the calculated quantity is
known as the propagated error. The propagated error can be estimated by
dy ≈ f ′(x)dx.
Δq
• To estimate the relative error of a particular quantity q, we estimate q .
4.3 Maxima and Minima
• A function may have both an absolute maximum and an absolute minimum, have just one absolute extremum, or
have no absolute maximum or absolute minimum.
• If a function has a local extremum, the point at which it occurs must be a critical point. However, a function need
not have a local extremum at a critical point.
• A continuous function over a closed, bounded interval has an absolute maximum and an absolute minimum. Each
extremum occurs at a critical point or an endpoint.
4.4 The Mean Value Theorem
• If f is continuous over [a, b] and differentiable over (a, b) and f (a) = 0 = f (b), then there exists a point
c ∈ (a, b) such that f ′ (c) = 0. This is Rolle’s theorem.
• If f is continuous over [a, b] and differentiable over (a, b), then there exists a point c ∈ (a, b) such that
f ′(c) =
f (b) − f (a)
.
b−a
502
Chapter 4 | Applications of Derivatives
This is the Mean Value Theorem.
• If f ′(x) = 0 over an interval I, then f is constant over I.
• If two differentiable functions f and g satisfy f ′(x) = g′(x) over I, then f (x) = g(x) + C for some constant
C.
• If f ′ (x) > 0 over an interval I, then f is increasing over I. If f ′(x) < 0 over I, then f is decreasing over
I.
4.5 Derivatives and the Shape of a Graph
• If c is a critical point of f and f ′ (x) > 0 for x < c and f ′ (x) < 0 for x > c, then f has a local maximum at
c.
• If c is a critical point of f and f ′ (x) < 0 for x < c and f ′ (x) > 0 for x > c, then f has a local minimum at
c.
• If f ″(x) > 0 over an interval I, then f is concave up over I.
• If f ″(x) < 0 over an interval I, then f is concave down over I.
• If f ′ (c) = 0 and f ″(c) > 0, then f has a local minimum at c.
• If f ′ (c) = 0 and f ″(c) < 0, then f has a local maximum at c.
• If f ′ (c) = 0 and f ″(c) = 0, then evaluate f ′ (x) at a test point x to the left of c and a test point x to the right
of c, to determine whether f has a local extremum at c.
4.6 Limits at Infinity and Asymptotes
• The limit of f (x) is L as x → ∞ (or as x → −∞) if the values f (x) become arbitrarily close to L as x
becomes sufficiently large.
• The limit of f (x) is ∞ as x → ∞ if f (x) becomes arbitrarily large as x becomes sufficiently large. The limit
of f (x) is −∞ as x → ∞ if f (x) < 0 and | f (x)| becomes arbitrarily large as x becomes sufficiently large. We
can define the limit of f (x) as x approaches −∞ similarly.
• For a polynomial function p(x) = a n x n + a n − 1 x n − 1 + … + a 1 x + a 0, where a n ≠ 0, the end behavior is
determined by the leading term a n x n. If n ≠ 0,
• For a rational function f (x) =
p(x) approaches ∞ or −∞ at each end.
p(x)
, the end behavior is determined by the relationship between the degree of p
q(x)
and the degree of q. If the degree of p is less than the degree of q, the line y = 0 is a horizontal asymptote for
f . If the degree of p is equal to the degree of q, then the line y =
an
is a horizontal asymptote, where a n and
bn
b n are the leading coefficients of p and q, respectively. If the degree of p is greater than the degree of q, then
f approaches ∞ or −∞ at each end.
4.7 Applied Optimization Problems
• To solve an optimization problem, begin by drawing a picture and introducing variables.
• Find an equation relating the variables.
• Find a function of one variable to describe the quantity that is to be minimized or maximized.
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Chapter 4 | Applications of Derivatives
503
• Look for critical points to locate local extrema.
4.8 L’Hôpital’s Rule
• L’Hôpital’s rule can be used to evaluate the limit of a quotient when the indeterminate form 0 or ∞/∞ arises.
0
• L’Hôpital’s rule can also be applied to other indeterminate forms if they can be rewritten in terms of a limit involving
a quotient that has the indeterminate form 0 or ∞/∞.
0
• The exponential function e x grows faster than any power function x p,
p > 0.
• The logarithmic function ln x grows more slowly than any power function x p,
p > 0.
4.9 Newton’s Method
• Newton’s method approximates roots of f (x) = 0 by starting with an initial approximation x 0, then uses tangent
lines to the graph of f to create a sequence of approximations x 1, x 2, x 3 ,….
• Typically, Newton’s method is an efficient method for finding a particular root. In certain cases, Newton’s method
fails to work because the list of numbers x 0, x 1, x 2 ,… does not approach a finite value or it approaches a value
other than the root sought.
• Any process in which a list of numbers x 0, x 1, x 2 ,… is generated by defining an initial number x 0 and defining
the subsequent numbers by the equation x n = F(x n − 1) for some function F is an iterative process. Newton’s
⎡ f (x) ⎤
for a given function f .
f ′ (x) ⎦
method is an example of an iterative process, where the function F(x) = x − ⎣
4.10 Antiderivatives
• If F is an antiderivative of f , then every antiderivative of f is of the form F(x) + C for some constant C.
• Solving the initial-value problem
dy
= f (x), y(x 0) = y 0
dx
requires us first to find the set of antiderivatives of f and then to look for the particular antiderivative that also
satisfies the initial condition.
CHAPTER 4 REVIEW EXERCISES
True or False? Justify your answer with a proof or a
counterexample. Assume that f (x) is continuous and
differentiable unless stated otherwise.
525. If f (−1) = −6 and f (1) = 2, then there exists at
least one point x ∈ [−1, 1] such that f ′ (x) = 4.
526. If f ′ (c) = 0, there is a maximum or minimum at
x = c.
527. There is a function such that f (x) < 0, f ′ (x) > 0,
and f ″(x) < 0. (A graphical “proof” is acceptable for this
answer.)
528. There is a function such that there is both an
inflection point and a critical point for some value x = a.
504
529.
Chapter 4 | Applications of Derivatives
Given the graph of f ′,
determine where f is
increasing or decreasing.
3x x 2 + 1
539. x lim
→∞
4
x −1
⎛ ⎞
540. x lim
cos⎝1x ⎠
→∞
541.
lim x − 1
x → 1 sin(πx)
(3x) 1/x
542. x lim
→∞
530. The graph of f is given below. Draw f ′.
Use Newton’s method to find the first two iterations, given
the starting point.
543. y = x 3 + 1, x 0 = 0.5
544.
1 = 1, x = 0
x+1 2 0
Find the antiderivatives F(x) of the following functions.
545. g(x) = x − 12
x
531.
Find
the
linear
approximation
L(x)
to
y = x 2 + tan(πx) near x = 1 .
4
532.
Find the differential of y = x 2 − 5x − 6 and
evaluate for x = 2 with dx = 0.1.
Find the critical points and the local and absolute extrema
of the following functions on the given interval.
533.
f (x) = x + sin 2 (x) over [0, π]
534.
f (x) = 3x 4 − 4x 3 − 12x 2 + 6 over [−3, 3]
Determine over which intervals the following functions are
increasing, decreasing, concave up, and concave down.
535. x(t) = 3t 4 − 8t 3 − 18t 2
536. y = x + sin(πx)
537. g(x) = x − x
538.
f (θ) = sin(3θ)
Evaluate the following limits.
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546.
f (x) = 2x + 6cos x, F(π) = π 2 + 2
Graph the following functions by hand. Make sure to label
the inflection points, critical points, zeros, and asymptotes.
547. y =
1
x(x + 1) 2
548. y = x − 4 − x 2
549. A car is being compacted into a rectangular solid.
The volume is decreasing at a rate of 2 m3/sec. The length
and width of the compactor are square, but the height is not
the same length as the length and width. If the length and
width walls move toward each other at a rate of 0.25 m/
sec, find the rate at which the height is changing when the
length and width are 2 m and the height is 1.5 m.
Chapter 4 | Applications of Derivatives
550. A rocket is launched into space; its kinetic energy
⎛ ⎞
is given by K(t) = ⎝1 ⎠m(t)v(t) 2, where K is the kinetic
2
energy in joules, m is the mass of the rocket in kilograms,
and v is the velocity of the rocket in meters/second.
Assume the velocity is increasing at a rate of 15 m/sec2
and the mass is decreasing at a rate of 10 kg/sec because
the fuel is being burned. At what rate is the rocket’s kinetic
energy changing when the mass is 2000 kg and the
velocity is 5000 m/sec? Give your answer in mega-Joules
(MJ), which is equivalent to 10 6 J.
551. The famous Regiomontanus’ problem for angle
maximization was proposed during the 15 th century. A
painting hangs on a wall with the bottom of the painting a
distance a feet above eye level, and the top b feet above
eye level. What distance x (in feet) from the wall should
the viewer stand to maximize the angle subtended by the
painting, θ ?
552. An airline sells tickets from Tokyo to Detroit for
$1200. There are 500 seats available and a typical flight
books 350 seats. For every $10 decrease in price, the
airline observes an additional five seats sold. What should
the fare be to maximize profit? How many passengers
would be onboard?
505
506
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Chapter 4 | Applications of Derivatives
Chapter 5 | Integration
507
5 | INTEGRATION
Figure 5.1 Iceboating is a popular winter sport in parts of the northern United States and Europe. (credit: modification of work
by Carter Brown, Flickr)
Chapter Outline
5.1 Approximating Areas
5.2 The Definite Integral
5.3 The Fundamental Theorem of Calculus
5.4 Integration Formulas and the Net Change Theorem
5.5 Substitution
5.6 Integrals Involving Exponential and Logarithmic Functions
5.7 Integrals Resulting in Inverse Trigonometric Functions
Introduction
Iceboats are a common sight on the lakes of Wisconsin and Minnesota on winter weekends. Iceboats are similar to sailboats,
but they are fitted with runners, or “skates,” and are designed to run over the ice, rather than on water. Iceboats can move
very quickly, and many ice boating enthusiasts are drawn to the sport because of the speed. Top iceboat racers can attain
508
Chapter 5 | Integration
speeds up to five times the wind speed. If we know how fast an iceboat is moving, we can use integration to determine how
far it travels. We revisit this question later in the chapter (see Example 5.27).
Determining distance from velocity is just one of many applications of integration. In fact, integrals are used in a wide
variety of mechanical and physical applications. In this chapter, we first introduce the theory behind integration and use
integrals to calculate areas. From there, we develop the Fundamental Theorem of Calculus, which relates differentiation and
integration. We then study some basic integration techniques and briefly examine some applications.
5.1 | Approximating Areas
Learning Objectives
5.1.1 Use sigma (summation) notation to calculate sums and powers of integers.
5.1.2 Use the sum of rectangular areas to approximate the area under a curve.
5.1.3 Use Riemann sums to approximate area.
Archimedes was fascinated with calculating the areas of various shapes—in other words, the amount of space enclosed by
the shape. He used a process that has come to be known as the method of exhaustion, which used smaller and smaller shapes,
the areas of which could be calculated exactly, to fill an irregular region and thereby obtain closer and closer approximations
to the total area. In this process, an area bounded by curves is filled with rectangles, triangles, and shapes with exact area
formulas. These areas are then summed to approximate the area of the curved region.
In this section, we develop techniques to approximate the area between a curve, defined by a function f (x), and the x-axis
on a closed interval ⎡⎣a, b⎤⎦. Like Archimedes, we first approximate the area under the curve using shapes of known area
(namely, rectangles). By using smaller and smaller rectangles, we get closer and closer approximations to the area. Taking
a limit allows us to calculate the exact area under the curve.
Let’s start by introducing some notation to make the calculations easier. We then consider the case when f (x) is continuous
and nonnegative. Later in the chapter, we relax some of these restrictions and develop techniques that apply in more general
cases.
Sigma (Summation) Notation
As mentioned, we will use shapes of known area to approximate the area of an irregular region bounded by curves. This
process often requires adding up long strings of numbers. To make it easier to write down these lengthy sums, we look at
some new notation here, called sigma notation (also known as summation notation). The Greek capital letter Σ, sigma,
is used to express long sums of values in a compact form. For example, if we want to add all the integers from 1 to 20
without sigma notation, we have to write
1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20.
We could probably skip writing a couple of terms and write
1 + 2 + 3 + 4 + ⋯ + 19 + 20,
which is better, but still cumbersome. With sigma notation, we write this sum as
20
∑ i,
i=1
which is much more compact.
Typically, sigma notation is presented in the form
n
∑ ai
i=1
where a i describes the terms to be added, and the i is called the index. Each term is evaluated, then we sum all the values,
beginning with the value when i = 1 and ending with the value when i = n. For example, an expression like
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7
∑ si
i=2
is
Chapter 5 | Integration
509
interpreted as s 2 + s 3 + s 4 + s 5 + s 6 + s 7. Note that the index is used only to keep track of the terms to be added; it does
not factor into the calculation of the sum itself. The index is therefore called a dummy variable. We can use any letter we
like for the index. Typically, mathematicians use i, j, k, m, and n for indices.
Let’s try a couple of examples of using sigma notation.
Example 5.1
Using Sigma Notation
a. Write in sigma notation and evaluate the sum of terms 3 i for i = 1, 2, 3, 4, 5.
b. Write the sum in sigma notation:
1+1 +1 + 1 + 1 .
4 9 16 25
Solution
a. Write
5
∑ 3i
i=1
= 3 + 32 + 33 + 34 + 35
= 363.
b. The denominator of each term is a perfect square. Using sigma notation, this sum can be written as
5
∑
i=1
5.1
1.
i2
Write in sigma notation and evaluate the sum of terms 2i for i = 3, 4, 5, 6.
The properties associated with the summation process are given in the following rule.
Rule: Properties of Sigma Notation
Let a 1, a 2 ,…, a n and b 1, b 2 ,…, b n represent two sequences of terms and let c be a constant. The following
properties hold for all positive integers n and for integers m, with 1 ≤ m ≤ n.
1.
n
∑ c = nc
(5.1)
n
n
(5.2)
i=1
i=1
i=1
2.
∑ ca i = c ∑ a i
3.
n
∑
i=1
⎛
⎝
a i + b i⎞⎠ =
n
n
i=1
i=1
n
n
i=1
i=1
∑ ai + ∑ bi
(5.3)
4.
n
∑
i=1
⎛
⎝
a i − b i⎞⎠ =
∑ ai − ∑ bi
(5.4)
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Chapter 5 | Integration
5.
n
m
n
i=1
i=1
i=m+1
∑ ai = ∑ ai + ∑
(5.5)
ai
Proof
We prove properties 2. and 3. here, and leave proof of the other properties to the Exercises.
2. We have
n
∑ ca i
i=1
= ca 1 + ca 2 + ca 3 + ⋯ + ca n
= c(a 1 + a 2 + a 3 + ⋯ + a n)
n
= c ∑ a i.
i=1
3. We have
n
∑
i=1
a i + b i⎞⎠ = ⎛⎝a 1 + b 1⎞⎠ + ⎛⎝a 2 + b 2⎞⎠ + ⎛⎝a 3 + b 3⎞⎠ + ⋯ + ⎛⎝a n + b n⎞⎠
⎛
⎝
= (a 1 + a 2 + a 3 + ⋯ + a n) + ⎛⎝b 1 + b 2 + b 3 + ⋯ + b n⎞⎠
=
n
n
i=1
i=1
∑ a i + ∑ b i.
□
A few more formulas for frequently found functions simplify the summation process further. These are shown in the next
rule, for sums and powers of integers, and we use them in the next set of examples.
Rule: Sums and Powers of Integers
1. The sum of n integers is given by
n
∑ i = 1 + 2 + ⋯ + n = n(n 2+ 1) .
i=1
2. The sum of consecutive integers squared is given by
n
+ 1)
∑ i 2 = 1 2 + 2 2 + ⋯ + n 2 = n(n + 1)(2n
.
6
i=1
3. The sum of consecutive integers cubed is given by
n
∑ i3 = 13 + 23 + ⋯ + n3 = n
i=1
Example 5.2
Evaluation Using Sigma Notation
Write using sigma notation and evaluate:
a. The sum of the terms (i − 3) 2 for i = 1, 2,…, 200.
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2
(n + 1) 2
.
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Chapter 5 | Integration
511
b. The sum of the terms ⎛⎝i 3 − i 2⎞⎠ for i = 1, 2, 3, 4, 5, 6.
Solution
a. Multiplying out (i − 3) 2, we can break the expression into three terms.
200
∑ (i − 3) 2
i=1
=
=
=
200
∑ ⎛⎝i 2 − 6i + 9⎞⎠
i=1
200
200
200
i=1
200
i=1
200
i=1
200
i=1
i=1
i=1
∑ i 2 − ∑ 6i + ∑ 9
∑ i2 − 6 ∑ i + ∑ 9
⎡200(200 + 1) ⎤
200(200 + 1)(400 + 1)
=
− 6⎣
⎦ + 9(200)
2
6
= 2,686,700 − 120,600 + 1800
= 2,567,900
b. Use sigma notation property iv. and the rules for the sum of squared terms and the sum of cubed terms.
6
∑ ⎛⎝i 3 − i 2⎞⎠
i=1
=
6
6
i=1
i=1
∑ i3 − ∑ i2
6 2 (6 + 1) 2 6(6 + 1)⎛⎝2(6) + 1⎞⎠
−
4
6
1764
546
=
−
4
6
= 350
=
5.2
Find the sum of the values of 4 + 3i for i = 1, 2,…, 100.
Example 5.3
Finding the Sum of the Function Values
Find the sum of the values of f (x) = x 3 over the integers 1, 2, 3,…, 10.
Solution
Using the formula, we have
10
∑ i3
i=0
=
(10) 2 (10 + 1) 2
4
100(121)
4
= 3025.
=
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Chapter 5 | Integration
5.3
Evaluate the sum indicated by the notation
20
∑
k=1
(2k + 1).
Approximating Area
Now that we have the necessary notation, we return to the problem at hand: approximating the area under a curve. Let f (x)
be a continuous, nonnegative function defined on the closed interval ⎡⎣a, b⎤⎦. We want to approximate the area A bounded by
f (x) above, the x-axis below, the line x = a on the left, and the line x = b on the right (Figure 5.2).
Figure 5.2 An area (shaded region) bounded by the curve
f (x) at top, the x-axis at bottom, the line x = a to the left, and
the line x = b at right.
How do we approximate the area under this curve? The approach is a geometric one. By dividing a region into many small
shapes that have known area formulas, we can sum these areas and obtain a reasonable estimate of the true area. We begin
a
by dividing the interval ⎡⎣a, b⎤⎦ into n subintervals of equal width, b −
n . We do this by selecting equally spaced points
x 0, x 1, x 2 ,…, x n with x 0 = a, x n = b, and
a
xi − xi − 1 = b −
n
for i = 1, 2, 3,…, n.
a
We denote the width of each subinterval with the notation Δx, so Δx = b −
n and
x i = x 0 + iΔx
for i = 1, 2, 3,…, n. This notion of dividing an interval ⎡⎣a, b⎤⎦ into subintervals by selecting points from within the interval
is used quite often in approximating the area under a curve, so let’s define some relevant terminology.
Definition
A set of points P = {x i} for i = 0, 1, 2,…, n with a = x 0 < x 1 < x 2 < ⋯ < x n = b, which divides the interval
⎡
⎤
into subintervals of the form [x 0, x 1], [x 1, x 2],…, [x n − 1, x n] is called a partition of ⎡⎣a, b⎤⎦. If the
⎣a, b⎦
subintervals all have the same width, the set of points forms a regular partition of the interval ⎡⎣a, b⎤⎦.
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Chapter 5 | Integration
513
We can use this regular partition as the basis of a method for estimating the area under the curve. We next examine two
methods: the left-endpoint approximation and the right-endpoint approximation.
Rule: Left-Endpoint Approximation
On each subinterval [x i − 1, x i] (for i = 1, 2, 3,…, n), construct a rectangle with width Δx and height equal to
f (x i − 1), which is the function value at the left endpoint of the subinterval. Then the area of this rectangle is
f (x i − 1)Δx. Adding the areas of all these rectangles, we get an approximate value for A (Figure 5.3). We use the
notation Ln to denote that this is a left-endpoint approximation of A using n subintervals.
A ≈ L n = f (x 0)Δx + f (x 1)Δx + ⋯ + f (x n − 1)Δx
=
n
∑
i=1
(5.6)
f (x i − 1)Δx
Figure 5.3 In the left-endpoint approximation of area under a
curve, the height of each rectangle is determined by the function
value at the left of each subinterval.
The second method for approximating area under a curve is the right-endpoint approximation. It is almost the same as the
left-endpoint approximation, but now the heights of the rectangles are determined by the function values at the right of each
subinterval.
Rule: Right-Endpoint Approximation
Construct a rectangle on each subinterval [x i − 1, x i], only this time the height of the rectangle is determined by the
function value f (x i) at the right endpoint of the subinterval. Then, the area of each rectangle is f (x i)Δx and the
approximation for A is given by
A ≈ R n = f (x 1)Δx + f (x 2)Δx + ⋯ + f (x n)Δx
=
n
∑
i=1
f (x i)Δx.
The notation R n indicates this is a right-endpoint approximation for A (Figure 5.4).
(5.7)
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Chapter 5 | Integration
Figure 5.4 In the right-endpoint approximation of area under
a curve, the height of each rectangle is determined by the
function value at the right of each subinterval. Note that the
right-endpoint approximation differs from the left-endpoint
approximation in Figure 5.3.
2
The graphs in Figure 5.5 represent the curve f (x) = x . In graph (a) we divide the region represented by the interval
2
[0, 3] into six subintervals, each of width 0.5. Thus, Δx = 0.5. We then form six rectangles by drawing vertical lines
perpendicular to x i − 1, the left endpoint of each subinterval. We determine the height of each rectangle by calculating
f (x i − 1) for i = 1, 2, 3, 4, 5, 6. The intervals are ⎡⎣0, 0.5⎤⎦, ⎡⎣0.5, 1⎤⎦, ⎡⎣1, 1.5⎤⎦, ⎡⎣1.5, 2⎤⎦, ⎡⎣2, 2.5⎤⎦, ⎡⎣2.5, 3⎤⎦. We find the area
of each rectangle by multiplying the height by the width. Then, the sum of the rectangular areas approximates the area
between f (x) and the x-axis. When the left endpoints are used to calculate height, we have a left-endpoint approximation.
Thus,
A ≈ L6 =
6
∑
i=1
f (x i − 1)Δx = f (x 0)Δx + f (x 1)Δx + f (x 2)Δx + f (x 3)Δx + f (x 4)Δx + f (x 5)Δx
= f (0)0.5 + f (0.5)0.5 + f (1)0.5 + f (1.5)0.5 + f (2)0.5 + f (2.5)0.5
= (0)0.5 + (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5
= 0 + 0.0625 + 0.25 + 0.5625 + 1 + 1.5625
= 3.4375.
Figure 5.5 Methods of approximating the area under a curve by using (a) the left endpoints
and (b) the right endpoints.
In Figure 5.5(b), we draw vertical lines perpendicular to x i such that x i is the right endpoint of each subinterval, and
calculate f (x i) for i = 1, 2, 3, 4, 5, 6. We multiply each f (x i) by Δx to find the rectangular areas, and then add them.
This is a right-endpoint approximation of the area under f (x). Thus,
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Chapter 5 | Integration
515
A ≈ R6 =
6
∑
i=1
f (x i)Δx = f (x 1)Δx + f (x 2)Δx + f (x 3)Δx + f (x 4)Δx + f (x 5)Δx + f (x 6)Δx
= f (0.5)0.5 + f (1)0.5 + f (1.5)0.5 + f (2)0.5 + f (2.5)0.5 + f (3)0.5
= (0.125)0.5 + (0.5)0.5 + (1.125)0.5 + (2)0.5 + (3.125)0.5 + (4.5)0.5
= 0.0625 + 0.25 + 0.5625 + 1 + 1.5625 + 2.25
= 5.6875.
Example 5.4
Approximating the Area Under a Curve
Use both left-endpoint and right-endpoint approximations to approximate the area under the curve of f (x) = x 2
on the interval [0, 2]; use n = 4.
Solution
First, divide the interval [0, 2] into n equal subintervals. Using n = 4, Δx =
(2 − 0)
= 0.5. This is the width of
4
each rectangle. The intervals ⎡⎣0, 0.5⎤⎦, ⎡⎣0.5, 1⎤⎦, ⎡⎣1, 1.5⎤⎦, ⎡⎣1.5, 2⎤⎦ are shown in Figure 5.6. Using a left-endpoint
approximation, the heights are f (0) = 0, f (0.5) = 0.25, f (1) = 1, f (1.5) = 2.25. Then,
L 4 = f (x 0)Δx + f (x 1)Δx + f (x 2)Δx + f (x 3)Δx
= 0(0.5) + 0.25(0.5) + 1(0.5) + 2.25(0.5)
= 1.75.
Figure 5.6 The graph shows the left-endpoint approximation
of the area under f (x) = x 2 from 0 to 2.
The right-endpoint approximation is shown in Figure 5.7. The intervals are the same, Δx = 0.5, but now use
the right endpoint to calculate the height of the rectangles. We have
R 4 = f (x 1)Δx + f (x 2)Δx + f (x 3)Δx + f (x 4)Δx
= 0.25(0.5) + 1(0.5) + 2.25(0.5) + 4(0.5)
= 3.75.
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Chapter 5 | Integration
Figure 5.7 The graph shows the right-endpoint approximation
of the area under f (x) = x 2 from 0 to 2.
The left-endpoint approximation is 1.75; the right-endpoint approximation is 3.75.
5.4
Sketch left-endpoint and right-endpoint approximations for
f (x) = 1x on [1, 2]; use n = 4.
Approximate the area using both methods.
Looking at Figure 5.5 and the graphs in Example 5.4, we can see that when we use a small number of intervals, neither
the left-endpoint approximation nor the right-endpoint approximation is a particularly accurate estimate of the area under
the curve. However, it seems logical that if we increase the number of points in our partition, our estimate of A will improve.
We will have more rectangles, but each rectangle will be thinner, so we will be able to fit the rectangles to the curve more
precisely.
We can demonstrate the improved approximation obtained through smaller intervals with an example. Let’s explore the idea
of increasing n, first in a left-endpoint approximation with four rectangles, then eight rectangles, and finally 32 rectangles.
Then, let’s do the same thing in a right-endpoint approximation, using the same sets of intervals, of the same curved region.
Figure 5.8 shows the area of the region under the curve f (x) = (x − 1) 3 + 4 on the interval [0, 2] using a left-endpoint
approximation where n = 4. The width of each rectangle is
Δx = 2 − 0 = 1 .
4
2
The area is approximated by the summed areas of the rectangles, or
L 4 = f (0)(0.5) + f (0.5)(0.5) + f (1)(0.5) + f (1.5)0.5
= 7.5.
Figure 5.8 With a left-endpoint approximation and dividing
the region from a to b into four equal intervals, the area under
the curve is approximately equal to the sum of the areas of the
rectangles.
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Chapter 5 | Integration
517
Figure 5.9 shows the same curve divided into eight subintervals. Comparing the graph with four rectangles in Figure 5.8
with this graph with eight rectangles, we can see there appears to be less white space under the curve when n = 8. This
white space is area under the curve we are unable to include using our approximation. The area of the rectangles is
L 8 = f (0)(0.25) + f (0.25)(0.25) + f (0.5)(0.25) + f (0.75)(0.25)
+ f (1)(0.25) + f (1.25)(0.25) + f (1.5)(0.25) + f (1.75)(0.25)
= 7.75.
Figure 5.9 The region under the curve is divided into n = 8
rectangular areas of equal width for a left-endpoint
approximation.
The graph in Figure 5.10 shows the same function with 32 rectangles inscribed under the curve. There appears to be little
white space left. The area occupied by the rectangles is
L 32 = f (0)(0.0625) + f (0.0625)(0.0625) + f (0.125)(0.0625) + ⋯ + f (1.9375)(0.0625)
= 7.9375.
Figure 5.10 Here, 32 rectangles are inscribed under the curve
for a left-endpoint approximation.
We can carry out a similar process for the right-endpoint approximation method. A right-endpoint approximation of the
same curve, using four rectangles (Figure 5.11), yields an area
R 4 = f (0.5)(0.5) + f (1)(0.5) + f (1.5)(0.5) + f (2)(0.5)
= 8.5.
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Chapter 5 | Integration
Figure 5.11 Now we divide the area under the curve into four
equal subintervals for a right-endpoint approximation.
Dividing the region over the interval [0, 2] into eight rectangles results in Δx = 2 − 0 = 0.25. The graph is shown in
8
Figure 5.12. The area is
R 8 = f (0.25)(0.25) + f (0.5)(0.25) + f (0.75)(0.25) + f (1)(0.25)
+ f (1.25)(0.25) + f (1.5)(0.25) + f (1.75)(0.25) + f (2)(0.25)
= 8.25.
Figure 5.12 Here we use right-endpoint approximation for a
region divided into eight equal subintervals.
Last, the right-endpoint approximation with n = 32 is close to the actual area (Figure 5.13). The area is approximately
R 32 = f (0.0625)(0.0625) + f (0.125)(0.0625) + f (0.1875)(0.0625) + ⋯ + f (2)(0.0625)
= 8.0625.
Figure 5.13 The region is divided into 32 equal subintervals
for a right-endpoint approximation.
Based on these figures and calculations, it appears we are on the right track; the rectangles appear to approximate the area
under the curve better as n gets larger. Furthermore, as n increases, both the left-endpoint and right-endpoint approximations
appear to approach an area of 8 square units. Table 5.1 shows a numerical comparison of the left- and right-endpoint
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Chapter 5 | Integration
519
methods. The idea that the approximations of the area under the curve get better and better as n gets larger and larger is very
important, and we now explore this idea in more detail.
Values of n
Approximate Area Ln
Approximate Area Rn
n=4
7.5
8.5
n=8
7.75
8.25
n = 32
7.94
8.06
Table 5.1 Converging Values of Left- and Right-Endpoint Approximations
as n Increases
Forming Riemann Sums
So far we have been using rectangles to approximate the area under a curve. The heights of these rectangles have been
determined by evaluating the function at either the right or left endpoints of the subinterval [x i − 1, x i]. In reality, there is
no reason to restrict evaluation of the function to one of these two points only. We could evaluate the function at any point
ci in the subinterval [x i − 1, x i], and use f ⎛⎝x*i ⎞⎠ as the height of our rectangle. This gives us an estimate for the area of
the form
A≈
n
∑
i=1
f ⎛⎝x*i ⎞⎠Δx.
A sum of this form is called a Riemann sum, named for the 19th-century mathematician Bernhard Riemann, who developed
the idea.
Definition
Let f (x) be defined on a closed interval ⎡⎣a, b⎤⎦ and let P be a regular partition of ⎡⎣a, b⎤⎦. Let Δx be the width of each
subinterval [x i − 1, x i] and for each i, let x*i be any point in [x i − 1, x i]. A Riemann sum is defined for f (x) as
n
∑
i=1
f ⎛⎝x*i ⎞⎠Δx.
Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as n get larger and
larger. The same thing happens with Riemann sums. Riemann sums give better approximations for larger values of n. We
are now ready to define the area under a curve in terms of Riemann sums.
Definition
Let f (x) be a continuous, nonnegative function on an interval ⎡⎣a, b⎤⎦, and let
f (x). Then, the area under the curve y = f (x) on ⎡⎣a, b⎤⎦ is given by
n
A = n lim
∑ f ⎛⎝x*i ⎞⎠Δx.
→∞
i=1
n
∑
i=1
f ⎛⎝x*i ⎞⎠Δx be a Riemann sum for
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Chapter 5 | Integration
See a graphical demonstration (http://www.openstaxcollege.org/l/20_riemannsums) of the
construction of a Riemann sum.
Some subtleties here are worth discussing. First, note that taking the limit of a sum is a little different from taking the limit
of a function f (x) as x goes to infinity. Limits of sums are discussed in detail in the chapter on Sequences and Series
(http://cnx.org/content/m53756/latest/) ; however, for now we can assume that the computational techniques we used
to compute limits of functions can also be used to calculate limits of sums.
Second, we must consider what to do if the expression converges to different limits for different choices of
⎧
⎨
⎩
⎫
x*i ⎬⎭.
Fortunately, this does not happen. Although the proof is beyond the scope of this text, it can be shown that if f (x) is
n
continuous on the closed interval ⎡⎣a, b⎤⎦, then n lim
∑ f ⎛⎝x*i ⎞⎠Δx exists and is unique (in other words, it does not depend
→∞
⎧
⎨
⎩
i=1
⎫
⎬
⎭
on the choice of x*i ).
⎧
⎫
We look at some examples shortly. But, before we do, let’s take a moment and talk about some specific choices for ⎨⎩x*i ⎬⎭.
⎧
Although any choice for ⎨⎩x*i
⎫
⎬
⎭
gives us an estimate of the area under the curve, we don’t necessarily know whether that
estimate is too high (overestimate) or too low (underestimate). If it is important to know whether our estimate is high or
⎧
low, we can select our value for ⎨⎩x*i
⎫
⎬
⎭
to guarantee one result or the other.
⎧
If we want an overestimate, for example, we can choose ⎨⎩x*i
⎧
x ∈ [x i − 1, x i]. In other words, we choose ⎨⎩x*i
⎧
the interval [x i − 1, x i]. If we select ⎨⎩x*i
⎫
⎬
⎭
⎫
⎬
⎭
⎫
⎬
⎭
such that for i = 1, 2, 3,…, n, f ⎛⎝x*i ⎞⎠ ≥ f (x) for all
so that for i = 1, 2, 3,…, n, f ⎛⎝x*i
n
in this way, then the Riemann sum
⎧
⎨
⎩
Similarly, if we want an underestimate, we can choose x*i
⎫
⎬
⎭
∑
i=1
⎞
⎠
is the maximum function value on
f ⎛⎝x*i ⎞⎠Δx is called an upper sum.
so that for i = 1, 2, 3,…, n, f ⎛⎝x*i
⎞
⎠
is the minimum function
value on the interval [x i − 1, x i]. In this case, the associated Riemann sum is called a lower sum. Note that if f (x) is either
increasing or decreasing throughout the interval ⎡⎣a, b⎤⎦, then the maximum and minimum values of the function occur at the
endpoints of the subintervals, so the upper and lower sums are just the same as the left- and right-endpoint approximations.
Example 5.5
Finding Lower and Upper Sums
Find a lower sum for f (x) = 10 − x 2 on [1, 2]; let n = 4 subintervals.
Solution
With
⎡
⎣
n=4
over
the
interval
[1, 2], Δx = 1 .
4
We
can
list
the
intervals
as
1, 1.25⎤⎦, ⎡⎣1.25, 1.5⎤⎦, ⎡⎣1.5, 1.75⎤⎦, ⎡⎣1.75, 2⎤⎦. Because the function is decreasing over the interval [1, 2], Figure
5.14 shows that a lower sum is obtained by using the right endpoints.
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Chapter 5 | Integration
521
Figure 5.14 The graph of f (x) = 10 − x 2 is set up for a
right-endpoint approximation of the area bounded by the curve
and the x-axis on [1, 2], and it shows a lower sum.
The Riemann sum is
4
∑
k=1
⎛
⎝10
− x 2⎞⎠(0.25) = 0.25⎡⎣10 − (1.25) 2 + 10 − (1.5) 2 + 10 − (1.75) 2 + 10 − (2) 2⎤⎦
= 0.25[8.4375 + 7.75 + 6.9375 + 6]
= 7.28.
The area of 7.28 is a lower sum and an underestimate.
5.5
a. Find an upper sum for f (x) = 10 − x 2 on [1, 2]; let n = 4.
b. Sketch the approximation.
Example 5.6
Finding Lower and Upper Sums for f(x) = sinx
⎡
⎤
Find a lower sum for f (x) = sin x over the interval ⎡⎣a, b⎤⎦ = ⎣0, π ⎦; let n = 6.
2
Solution
Let’s first look at the graph in Figure 5.15 to get a better idea of the area of interest.
522
Chapter 5 | Integration
Figure 5.15 The graph of y = sin x is divided into six regions: Δx = π/2 = π .
6
⎤ ⎡
⎤ ⎡
⎤ ⎡
⎤
⎡
⎤ ⎡
The intervals are ⎣0, π ⎦, ⎣ π , π ⎦, ⎣ π , π ⎦, ⎣ π , π ⎦, ⎣ π , 5π ⎦,
12 12 6 6 4 4 3 3 12
12
⎡
⎤
and ⎣ 5π , π ⎦. Note that f (x) = sin x is
12 2
⎡
⎤
increasing on the interval ⎣0, π ⎦, so a left-endpoint approximation gives us the lower sum. A left-endpoint
2
approximation is the Riemann sum
5
π ⎞.
∑ sin x i ⎛⎝12
⎠
i=0
We have
⎛ ⎞⎛ ⎞
⎛ ⎞⎛ ⎞
⎛ ⎞
⎛ ⎞⎛ ⎞
⎛ ⎞⎛ ⎞
⎛ ⎞⎛ ⎞
A ≈ sin(0)⎝ π ⎠ + sin⎝ π ⎠⎝ π ⎠ + sin⎝π ⎠⎝ π ⎠ + sin⎝π ⎠⎝ π ⎠ + sin⎝π ⎠⎝ π ⎠ + sin⎝5π ⎠⎝ π ⎠
3 12
12 12
12
12 12
4 12
6 12
= 0.863.
5.6
⎡
⎤
Using the function f (x) = sin x over the interval ⎣0, π ⎦, find an upper sum; let n = 6.
2
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Chapter 5 | Integration
523
5.1 EXERCISES
1. State whether the given sums are equal or unequal.
a.
b.
c.
d.
10
∑i
i=1
10
∑i
i=1
and
and
10
∑
k=1
k
15
∑ (i − 5)
11.
i=6
10
∑ i(i − 1)
i=1
10
∑ i(i − 1)
i=1
and
and
9
∑
j=0
10
∑
k=1
⎛
⎝
⎛ 2
⎝k
3.
− k⎞⎠
∑i
10
∑ i2
i=5
100
∑ a i = 15
i=1
and
100
∑ b i = −12.
i=1
In the
following exercises, compute the sums.
5.
6.
7.
100
i=1
100
∑
i=1
100
∑
i=1
100
∑
i=1
⎛
⎝
a i + b i⎞⎠
⎛
⎝
8.
9.
∑
k=1
50
∑
j=1
k=1
− 10 j⎞⎠
⎡
2
⎣(2k)
− 100k⎤⎦
12. L4 for f (x) =
1 on [2, 3]
x−1
ai − b
15. R6 for f (x) =
1
on ⎡⎣2, 5⎤⎦
x(x − 1)
16. R4 for
1
on [−2, 2]
x2 + 1
17. L4 for
1
on [−2, 2]
x2 + 1
20. Compute the left and right Riemann sums—L4 and R4,
respectively—for f (x) = (2 − |x|) on [−2, 2]. Compute
3a i − 4b i⎞⎠
⎛
⎝
their average value and compare it with the area under the
graph of f.
21. Compute the left and right Riemann sums—L6 and
R6, respectively—for f (x) = (3 − |3 − x|) on ⎡⎣0, 6⎤⎦.
5a i + 4b i⎞⎠
⎛
⎝
− 5k +
1
on ⎡⎣2, 5⎤⎦
x(x − 1)
19. L8 for x 2 − 2x + 1 on [0, 2]
⎞
i⎠
100⎛⎝k 2
14. L6 for f (x) =
18. R4 for x 2 − 2x + 1 on [0, 2]
In the following exercises, use summation properties and
formulas to rewrite and evaluate the sums.
20
25
∑
⎛ 2
⎝j
13. R4 for g(x) = cos(πx) on [0, 1]
i=5
∑
j = 11
In the following exercises, compute the indicated left and
right sums for the given functions on the indicated interval.
10
Suppose that
4.
20
∑
Let L n denote the left-endpoint sum using n subintervals
and let R n denote the corresponding right-endpoint sum.
j + 1⎞⎠ j
In the following exercises, use the rules for sums of powers
of integers to compute the sums.
2.
10.
1⎞⎠
Compute their average value and compare it with the area
under the graph of f.
22. Compute the left and right Riemann sums—L4 and
R4, respectively—for f (x) = 4 − x 2 on [−2, 2] and
compare their values.
23. Compute the left and right Riemann sums—L6 and
⎛ 2
⎝j
− 2 j⎞⎠
R6, respectively—for f (x) = 9 − (x − 3) 2 on ⎡⎣0, 6⎤⎦ and
compare their values.
Express the following endpoint sums in sigma notation but
do not evaluate them.
524
Chapter 5 | Integration
24. L30 for f (x) = x 2 on [1, 2]
37. To help get in shape, Joe gets a new pair of running
shoes. If Joe runs 1 mi each day in week 1 and adds 1 mi
25. L10 for f (x) = 4 − x 2 on [−2, 2]
26. R20 for f (x) = sin x on [0, π]
27. R100 for ln x on [1, e]
In the following exercises, graph the function then use a
calculator or a computer program to evaluate the following
left and right endpoint sums. Is the area under the curve
between the left and right endpoint sums?
28. [T] L100 and R100 for y = x 2 − 3x + 1 on the interval
[−1, 1]
29. [T] L100 and R100 for y = x 2 on the interval [0, 1]
30. [T] L50 and R50 for y = x2+ 1 on the interval [2, 4]
x −1
31. [T] L100 and R100 for y = x 3 on the interval [−1, 1]
⎡
⎤
32. [T] L50 and R50 for y = tan(x) on the interval ⎣0, π ⎦
4
33. [T] L100 and R100 for y = e 2x on the interval [−1, 1]
34. Let tj denote the time that it took Tejay van Garteren
to ride the jth stage of the Tour de France in 2014. If there
were a total of 21 stages, interpret
21
∑
j=1
t j.
35. Let r j denote the total rainfall in Portland on the jth
day of the year in 2009. Interpret
31
∑
j=1
r j.
36. Let d j denote the hours of daylight and δ j denote the
increase in the hours of daylight from day j − 1 to day j
in Fargo, North Dakota, on the jth day of the year. Interpret
d1 +
365
∑
j=2
δ j.
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10
to his daily routine each week, what is the total mileage on
Joe’s shoes after 25 weeks?
38. The following table gives approximate values of the
average annual atmospheric rate of increase in carbon
dioxide (CO2) each decade since 1960, in parts per million
(ppm). Estimate the total increase in atmospheric CO2
between 1964 and 2013.
Decade
Ppm/y
1964–1973
1.07
1974–1983
1.34
1984–1993
1.40
1994–2003
1.87
2004–2013
2.07
Table 5.2 Average Annual
Atmospheric CO2
Increase,
1964–2013 Source:
http://www.esrl.noaa.gov/
gmd/ccgg/trends/.
Chapter 5 | Integration
525
39. The following table gives the approximate increase in
sea level in inches over 20 years starting in the given year.
Estimate the net change in mean sea level from 1870 to
2010.
Starting Year
1870
1890
1910
1930
1950
1970
1990
20-Year Change
40. The following table gives the approximate increase in
dollars in the average price of a gallon of gas per decade
since 1950. If the average price of a gallon of gas in 2010
was $2.60, what was the average price of a gallon of gas in
1950?
Starting Year
10-Year Change
1950
0.03
1960
0.05
1970
0.86
1980
−0.03
1990
0.29
2000
1.12
0.3
1.5
0.2
2.8
0.7
1.1
1.5
Table 5.3 Approximate 20-Year Sea
Level Increases, 1870–1990 Source:
http://link.springer.com/article/
10.1007%2Fs10712-011-9119-1
Table 5.4 Approximate 10-Year Gas
Price Increases, 1950–2000 Source:
http://epb.lbl.gov/homepages/
Rick_Diamond/docs/
lbnl55011-trends.pdf.
526
Chapter 5 | Integration
41. The following table gives the percent growth of the
U.S. population beginning in July of the year indicated. If
the U.S. population was 281,421,906 in July 2000, estimate
the U.S. population in July 2010.
Year
% Change/Year
2000
1.12
2001
0.99
2002
0.93
43.
44.
2003
0.86
2004
0.93
2005
0.93
2006
0.97
2007
0.96
2008
0.95
2009
0.88
45.
Table 5.5 Annual Percentage
Growth of U.S. Population,
2000–2009 Source:
http://www.census.gov/
popest/data.
(Hint: To obtain the population in July 2001, multiply the
population in July 2000 by 1.0112 to get 284,573,831.)
In the following exercises, estimate the areas under the
curves by computing the left Riemann sums, L8.
42.
46. [T] Use a computer algebra system to compute the
Riemann sum,
L N,
for
N = 10, 30, 50 for
f (x) = 1 − x 2 on [−1, 1].
47. [T] Use a computer algebra system to compute the
Riemann
sum,
LN ,
for
N = 10, 30, 50
for
f (x) =
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1
on [−1, 1].
1 + x2
Chapter 5 | Integration
527
48. [T] Use a computer algebra system to compute the
Riemann sum, LN, for N = 10, 30, 50 for f (x) = sin 2 x
on [0, 2π]. Compare these estimates with π.
In the following exercises, use a calculator or a computer
program to evaluate the endpoint sums RN and LN for
N = 1,10,100. How do these estimates compare with the
57. For each of the three graphs:
a. Obtain a lower bound L(A) for the area enclosed
by the curve by adding the areas of the squares
enclosed completely by the curve.
b. Obtain an upper bound U(A) for the area by
adding to L(A) the areas B(A) of the squares
enclosed partially by the curve.
exact answers, which you can find via geometry?
49. [T] y = cos(πx) on the interval [0, 1]
50. [T] y = 3x + 2 on the interval ⎡⎣3, 5⎤⎦
In the following exercises, use a calculator or a computer
program to evaluate the endpoint sums RN and LN for
N = 1,10,100.
[T] y = x 4 − 5x 2 + 4 on the interval [−2, 2],
51.
which has an exact area of 32
15
52. [T] y = ln x on the interval [1, 2], which has an
exact area of 2ln(2) − 1
Explain why, if f (a) ≥ 0 and f is increasing on
53.
⎡
⎣
⎤
⎦
a, b , that the left endpoint estimate is a lower bound for
the area below the graph of f on ⎡⎣a, b⎤⎦.
54.
⎡
⎣
Explain why, if f (b) ≥ 0 and f is decreasing on
⎤
⎦
a, b , that the left endpoint estimate is an upper bound for
the area below the graph of f on ⎡⎣a, b⎤⎦.
55.
Show
that,
f (b) − f (a)
R N − L N = (b − a) ×
.
N
in
general,
56. Explain why, if f is increasing on ⎡⎣a, b⎤⎦, the error
between either LN or RN and the area A below the graph of
f is at most (b − a)
f (b) − f (a)
.
N
58. In the previous exercise, explain why L(A) gets no
smaller while U(A) gets no larger as the squares are
subdivided into four boxes of equal area.
528
59. A unit circle is made up of n wedges equivalent to the
inner wedge in the figure. The base of the inner triangle
⎞
is 1 unit and its height is sin⎛⎝ π
n ⎠. The base of the outer
⎞
⎛π ⎞
⎛π ⎞
triangle is B = cos⎛⎝ π
n ⎠ + sin⎝ n ⎠tan⎝ n ⎠ and the height is
⎛ ⎞
H = Bsin⎝2π
n ⎠. Use this information to argue that the area
of a unit circle is equal to π.
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Chapter 5 | Integration
Chapter 5 | Integration
529
5.2 | The Definite Integral
Learning Objectives
5.2.1
5.2.2
5.2.3
5.2.4
5.2.5
5.2.6
State the definition of the definite integral.
Explain the terms integrand, limits of integration, and variable of integration.
Explain when a function is integrable.
Describe the relationship between the definite integral and net area.
Use geometry and the properties of definite integrals to evaluate them.
Calculate the average value of a function.
In the preceding section we defined the area under a curve in terms of Riemann sums:
n
A = n lim
∑ f ⎛⎝x*i ⎞⎠Δx.
→∞
i=1
However, this definition came with restrictions. We required f (x) to be continuous and nonnegative. Unfortunately, realworld problems don’t always meet these restrictions. In this section, we look at how to apply the concept of the area under
the curve to a broader set of functions through the use of the definite integral.
Definition and Notation
The definite integral generalizes the concept of the area under a curve. We lift the requirements that f (x) be continuous
and nonnegative, and define the definite integral as follows.
Definition
If f (x) is a function defined on an interval ⎡⎣a, b⎤⎦, the definite integral of f from a to b is given by
n
b
∑
∫ a f (x)dx = n lim
→∞
i=1
f ⎛⎝x*i ⎞⎠Δx,
(5.8)
provided the limit exists. If this limit exists, the function f (x) is said to be integrable on ⎡⎣a, b⎤⎦, or is an integrable
function.
The integral symbol in the previous definition should look familiar. We have seen similar notation in the chapter on
Applications of Derivatives, where we used the indefinite integral symbol (without the a and b above and below) to
represent an antiderivative. Although the notation for indefinite integrals may look similar to the notation for a definite
integral, they are not the same. A definite integral is a number. An indefinite integral is a family of functions. Later in this
chapter we examine how these concepts are related. However, close attention should always be paid to notation so we know
whether we’re working with a definite integral or an indefinite integral.
Integral notation goes back to the late seventeenth century and is one of the contributions of Gottfried Wilhelm Leibniz, who
is often considered to be the codiscoverer of calculus, along with Isaac Newton. The integration symbol ∫ is an elongated S,
suggesting sigma or summation. On a definite integral, above and below the summation symbol are the boundaries of the
interval, ⎡⎣a, b⎤⎦. The numbers a and b are x-values and are called the limits of integration; specifically, a is the lower limit
and b is the upper limit. To clarify, we are using the word limit in two different ways in the context of the definite integral.
First, we talk about the limit of a sum as n → ∞. Second, the boundaries of the region are called the limits of integration.
We call the function f (x) the integrand, and the dx indicates that f (x) is a function with respect to x, called the variable
of integration. Note that, like the index in a sum, the variable of integration is a dummy variable, and has no impact on the
computation of the integral. We could use any variable we like as the variable of integration:
b
b
b
∫ a f (x)dx = ∫ a f (t)dt = ∫ a f (u)du
530
Chapter 5 | Integration
n
Previously, we discussed the fact that if f (x) is continuous on ⎡⎣a, b⎤⎦, then the limit n lim
∑ f ⎛⎝x*i ⎞⎠Δx exists and is
→∞
i=1
unique. This leads to the following theorem, which we state without proof.
Theorem 5.1: Continuous Functions Are Integrable
If f (x) is continuous on ⎡⎣a, b⎤⎦, then f is integrable on ⎡⎣a, b⎤⎦.
Functions that are not continuous on ⎡⎣a, b⎤⎦ may still be integrable, depending on the nature of the discontinuities. For
example, functions with a finite number of jump discontinuities on a closed interval are integrable.
It is also worth noting here that we have retained the use of a regular partition in the Riemann sums. This restriction is not
strictly necessary. Any partition can be used to form a Riemann sum. However, if a nonregular partition is used to define
the definite integral, it is not sufficient to take the limit as the number of subintervals goes to infinity. Instead, we must take
the limit as the width of the largest subinterval goes to zero. This introduces a little more complex notation in our limits and
makes the calculations more difficult without really gaining much additional insight, so we stick with regular partitions for
the Riemann sums.
Example 5.7
Evaluating an Integral Using the Definition
Use the definition of the definite integral to evaluate
∫
2
0
x 2 dx. Use a right-endpoint approximation to generate
the Riemann sum.
Solution
We first want to set up a Riemann sum. Based on the limits of integration, we have a = 0 and b = 2. For
i = 0, 1, 2,…, n, let P = {x i} be a regular partition of [0, 2]. Then
a 2
Δx = b −
n = n.
Since we are using a right-endpoint approximation to generate Riemann sums, for each i, we need to calculate
the function value at the right endpoint of the interval [x i − 1, x i]. The right endpoint of the interval is x i, and
since P is a regular partition,
⎡ ⎤
x i = x 0 + iΔx = 0 + i⎣ 2n ⎦ = 2i
n.
Thus, the function value at the right endpoint of the interval is
2
⎞
4i 2
f (x i) = x i2 = ⎛⎝2i
n⎠ = 2.
n
Then the Riemann sum takes the form
n
∑
i=1
f (x i)Δx =
∑ ⎛⎝4i2 ⎞⎠2n = ∑
n
i=1
2
n
n
Using the summation formula for
∑ i 2,
i=1
we have
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n
i=1
n
8i 2 = 8 ∑ i 2.
n3
n3i = 1
Chapter 5 | Integration
531
n
∑
i=1
n
f (x i)Δx = 83 ∑ i 2
n i=1
⎡n(n + 1)(2n + 1) ⎤
= 83 ⎣
⎦
6
n
2
⎡ 3
⎤
= 83 ⎣2n + 3n + n ⎦
6
n
3
2
+n
= 16n + 24n
3
6n
= 8 + 4n + 1 2 .
3
6n
Now, to calculate the definite integral, we need to take the limit as n → ∞. We get
∫
2
0
n
x 2 dx = n lim
∑ f (x i)Δx
→∞
i=1
⎛8 4
⎞
= n lim
+ n + 12
→ ∞⎝3
6n ⎠
⎛ ⎞
⎛8 ⎞
= n lim
+ lim ⎛4 ⎞ + lim 1
→ ∞⎝3 ⎠ n → ∞⎝n ⎠ n → ∞⎝6n 2 ⎠
= 8 + 0 + 0 = 8.
3
3
5.7
Use the definition of the definite integral to evaluate
∫
3
0
(2x − 1)dx. Use a right-endpoint approximation
to generate the Riemann sum.
Evaluating Definite Integrals
Evaluating definite integrals this way can be quite tedious because of the complexity of the calculations. Later in this chapter
we develop techniques for evaluating definite integrals without taking limits of Riemann sums. However, for now, we can
rely on the fact that definite integrals represent the area under the curve, and we can evaluate definite integrals by using
geometric formulas to calculate that area. We do this to confirm that definite integrals do, indeed, represent areas, so we can
then discuss what to do in the case of a curve of a function dropping below the x-axis.
Example 5.8
Using Geometric Formulas to Calculate Definite Integrals
Use the formula for the area of a circle to evaluate
∫
6
3
9 − (x − 3) 2dx.
Solution
The function describes a semicircle with radius 3. To find
532
Chapter 5 | Integration
∫
6
3
9 − (x − 3) 2dx,
we want to find the area under the curve over the interval ⎡⎣3, 6⎤⎦. The formula for the area of a circle is A = πr 2.
⎛ ⎞
The area of a semicircle is just one-half the area of a circle, or A = ⎝1 ⎠πr 2. The shaded area in Figure 5.16
2
⎛ ⎞
covers one-half of the semicircle, or A = ⎝1 ⎠πr 2. Thus,
4
∫
6
3
9 − (x − 3) 2 = 1 π(3) 2
4
= 9π
4
≈ 7.069.
Figure 5.16 The value of the integral of the function f (x)
over the interval ⎡⎣3, 6⎤⎦ is the area of the shaded region.
5.8
Use the formula for the area of a trapezoid to evaluate
∫
4
2
(2x + 3)dx.
Area and the Definite Integral
When we defined the definite integral, we lifted the requirement that f (x) be nonnegative. But how do we interpret “the
area under the curve” when f (x) is negative?
Net Signed Area
Let us return to the Riemann sum. Consider, for example, the function f (x) = 2 − 2x 2 (shown in Figure 5.17) on
⎧
the interval [0, 2]. Use n = 8 and choose ⎨⎩x*i } as the left endpoint of each interval. Construct a rectangle on each
subinterval of height f ⎛⎝x*i
⎞
⎠
and width Δx. When f ⎛⎝x*i
rectangle, as before. When f ⎛⎝x*i
⎞
⎠
⎞
⎠
is positive, the product f ⎛⎝x*i ⎞⎠Δx represents the area of the
is negative, however, the product f ⎛⎝x*i ⎞⎠Δx represents the negative of the area of the
rectangle. The Riemann sum then becomes
8
∑
i=1
f ⎛⎝x*i ⎞⎠Δx = ⎛⎝Area of rectangles above the x-axis⎞⎠ − ⎛⎝Area of rectangles below the x-axis⎞⎠
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Chapter 5 | Integration
533
Figure 5.17 For a function that is partly negative, the
Riemann sum is the area of the rectangles above the x-axis less
the area of the rectangles below the x-axis.
Taking the limit as n → ∞, the Riemann sum approaches the area between the curve above the x-axis and the x-axis, less
the area between the curve below the x-axis and the x-axis, as shown in Figure 5.18. Then,
∫
2
0
n
f (x)dx = n lim
∑ f (c i)Δx
→∞
i=1
= A 1 − A 2.
The quantity A 1 − A 2 is called the net signed area.
Figure 5.18 In the limit, the definite integral equals area A1
less area A2, or the net signed area.
Notice that net signed area can be positive, negative, or zero. If the area above the x-axis is larger, the net signed area is
positive. If the area below the x-axis is larger, the net signed area is negative. If the areas above and below the x-axis are
equal, the net signed area is zero.
Example 5.9
Finding the Net Signed Area
534
Chapter 5 | Integration
Find the net signed area between the curve of the function f (x) = 2x and the x-axis over the interval [−3, 3].
Solution
The function produces a straight line that forms two triangles: one from x = −3 to x = 0 and the other from
x = 0 to x = 3 (Figure 5.19). Using the geometric formula for the area of a triangle, A = 1 bh, the area of
2
triangle A1, above the axis, is
A 1 = 1 3(6) = 9,
2
where 3 is the base and 2(3) = 6 is the height. The area of triangle A2, below the axis, is
A 2 = 1 (3)(6) = 9,
2
where 3 is the base and 6 is the height. Thus, the net area is
∫
3
2xdx = A 1 − A 2 = 9 − 9 = 0.
−3
Figure 5.19 The area above the curve and below the x-axis
equals the area below the curve and above the x-axis.
Analysis
If A1 is the area above the x-axis and A2 is the area below the x-axis, then the net area is A 1 − A 2. Since the areas
of the two triangles are equal, the net area is zero.
5.9
Find the net signed area of f (x) = x − 2 over the interval ⎡⎣0, 6⎤⎦, illustrated in the following image.
Total Area
One application of the definite integral is finding displacement when given a velocity function. If v(t) represents the
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Chapter 5 | Integration
535
velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original
position. This is a very important application of the definite integral, and we examine it in more detail later in the chapter.
For now, we’re just going to look at some basics to get a feel for how this works by studying constant velocities.
When velocity is a constant, the area under the curve is just velocity times time. This idea is already very familiar. If a car
travels away from its starting position in a straight line at a speed of 75 mph for 2 hours, then it is 150 mi away from its
original position (Figure 5.20). Using integral notation, we have
∫
2
0
75dt = 150.
Figure 5.20 The area under the curve v(t) = 75 tells us how far the car
is from its starting point at a given time.
In the context of displacement, net signed area allows us to take direction into account. If a car travels straight north at a
speed of 60 mph for 2 hours, it is 120 mi north of its starting position. If the car then turns around and travels south at a
speed of 40 mph for 3 hours, it will be back at it starting position (Figure 5.21). Again, using integral notation, we have
∫
In this case the displacement is zero.
2
0
5
60dt + ∫ −40dt = 120 − 120
2
= 0.
536
Chapter 5 | Integration
Figure 5.21 The area above the axis and the area below the axis
are equal, so the net signed area is zero.
Suppose we want to know how far the car travels overall, regardless of direction. In this case, we want to know the area
between the curve and the x-axis, regardless of whether that area is above or below the axis. This is called the total area.
Graphically, it is easiest to think of calculating total area by adding the areas above the axis and the areas below the axis
(rather than subtracting the areas below the axis, as we did with net signed area). To accomplish this mathematically, we use
the absolute value function. Thus, the total distance traveled by the car is
2
∫ |60|dt + ∫
0
5
2
2
5
|−40|dt = ∫ 60dt + ∫ 40dt
0
= 120 + 120
= 240.
2
Bringing these ideas together formally, we state the following definitions.
Definition
Let f (x) be an integrable function defined on an interval ⎡⎣a, b⎤⎦. Let A1 represent the area between f (x) and the
x-axis that lies above the axis and let A2 represent the area between f (x) and the x-axis that lies below the axis. Then,
the net signed area between f (x) and the x-axis is given by
b
∫ a f (x)dx = A 1 − A 2.
The total area between f (x) and the x-axis is given by
b
∫ a| f (x)|dx = A 1 + A 2.
Example 5.10
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Chapter 5 | Integration
537
Finding the Total Area
Find the total area between f (x) = x − 2 and the x-axis over the interval ⎡⎣0, 6⎤⎦.
Solution
Calculate the x-intercept as (2, 0) (set y = 0, solve for x). To find the total area, take the area below the x-axis
over the subinterval [0, 2] and add it to the area above the x-axis on the subinterval ⎡⎣2, 6⎤⎦ (Figure 5.22).
Figure 5.22 The total area between the line and the x-axis
over ⎡⎣0, 6⎤⎦ is A2 plus A1.
We have
6
∫ |(x − 2)|dx = A 2 + A 1.
0
Then, using the formula for the area of a triangle, we obtain
A 2 = 1 bh = 1 · 2 · 2 = 2
2
2
1
1
A 1 = bh = · 4 · 4 = 8.
2
2
The total area, then, is
A 1 + A 2 = 8 + 2 = 10.
5.10
Find the total area between the function f (x) = 2x and the x-axis over the interval [−3, 3].
Properties of the Definite Integral
The properties of indefinite integrals apply to definite integrals as well. Definite integrals also have properties that relate to
the limits of integration. These properties, along with the rules of integration that we examine later in this chapter, help us
manipulate expressions to evaluate definite integrals.
Rule: Properties of the Definite Integral
1.
a
∫ a f (x)dx = 0
(5.9)
538
Chapter 5 | Integration
If the limits of integration are the same, the integral is just a line and contains no area.
2.
∫
a
b
b
(5.10)
f (x)dx = −∫ f (x)dx
a
If the limits are reversed, then place a negative sign in front of the integral.
3.
b
∫a
b
b
f (x) + g(x)⎤⎦dx = ∫ f (x)dx + ∫ g(x)dx
⎡
⎣
(5.11)
a
a
The integral of a sum is the sum of the integrals.
4.
⌠ f (x) − g(x) dx = ⌠ f (x)dx − ∫ bg(x)dx
a
⌡a
⌡a
b
⎡
⎣
b
⎤
⎦
(5.12)
The integral of a difference is the difference of the integrals.
5.
b
b
(5.13)
∫ a c f (x)dx = c∫ a f (x)
for constant c. The integral of the product of a constant and a function is equal to the constant multiplied by
the integral of the function.
6.
b
c
b
∫ a f (x)dx = ∫ a f (x)dx + ∫ c
f (x)dx
(5.14)
Although this formula normally applies when c is between a and b, the formula holds for all values of a, b, and
c, provided f (x) is integrable on the largest interval.
Example 5.11
Using the Properties of the Definite Integral
Use the properties of the definite integral to express the definite integral of f (x) = −3x 3 + 2x + 2 over the
interval [−2, 1] as the sum of three definite integrals.
Solution
Using integral notation, we have
∫
1
⎛
3
⎝−3x
−2
+ 2x + 2⎞⎠dx. We apply properties 3. and 5. to get
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Chapter 5 | Integration
539
∫
1
⎛
3
⎝−3x
−2
+ 2x + 2⎞⎠dx = ∫
1
−3x 3 dx + ∫
−2
= −3∫
5.11
1
−2
1
2xdx + ∫
−2
1
x 3 dx + 2∫
−2
xdx + ∫
1
2dx
−2
1
2dx.
−2
Use the properties of the definite integral to express the definite integral of f (x) = 6x 3 − 4x 2 + 2x − 3
over the interval [1, 3] as the sum of four definite integrals.
Example 5.12
Using the Properties of the Definite Integral
If it is known that
∫
8
0
f (x)dx = 10 and
∫
5
0
f (x)dx = 5, find the value of
∫
8
5
f (x)dx.
Solution
By property 6.,
b
c
b
∫ a f (x)dx = ∫ a f (x)dx + ∫ c
f (x)dx.
Thus,
∫
8
0
f (x)dx =
∫
5
0
8
f (x)dx + ∫ f (x)dx
8
5
10 = 5 + ∫ f (x)dx
5.12
If it is known that
∫
5
1
5 =
∫
f (x)dx = −3 and
∫
8
5
5
2
5
f (x)dx.
f (x)dx = 4, find the value of
∫
2
1
f (x)dx.
Comparison Properties of Integrals
A picture can sometimes tell us more about a function than the results of computations. Comparing functions by their graphs
as well as by their algebraic expressions can often give new insight into the process of integration. Intuitively, we might say
that if a function f (x) is above another function g(x), then the area between f (x) and the x-axis is greater than the area
between g(x) and the x-axis. This is true depending on the interval over which the comparison is made. The properties of
definite integrals are valid whether a < b, a = b, or a > b. The following properties, however, concern only the case
a ≤ b, and are used when we want to compare the sizes of integrals.
540
Chapter 5 | Integration
Theorem 5.2: Comparison Theorem
i. If f (x) ≥ 0 for a ≤ x ≤ b, then
b
∫ a f (x)dx ≥ 0.
ii. If f (x) ≥ g(x) for a ≤ x ≤ b, then
b
b
∫ a f (x)dx ≥ ∫ a g(x)dx.
iii. If m and M are constants such that m ≤ f (x) ≤ M for a ≤ x ≤ b, then
b
m(b − a) ≤ ∫ f (x)dx
a
≤ M(b − a).
Example 5.13
Comparing Two Functions over a Given Interval
Compare f (x) = 1 + x 2 and g(x) = 1 + x over the interval [0, 1].
Solution
Graphing these functions is necessary to understand how they compare over the interval [0, 1]. Initially, when
graphed on a graphing calculator, f (x) appears to be above g(x) everywhere. However, on the interval [0, 1],
the graphs appear to be on top of each other. We need to zoom in to see that, on the interval [0, 1], g(x) is above
f (x). The two functions intersect at x = 0 and x = 1 (Figure 5.23).
Figure 5.23 (a) The function f (x) appears above the function g(x)
except over the interval [0, 1] (b) Viewing the same graph with a greater
zoom shows this more clearly.
We can see from the graph that over the interval [0, 1], g(x) ≥ f (x). Comparing the integrals over the specified
interval [0, 1], we also see that
∫
1
0
1
g(x)dx ≥ ∫ f (x)dx (Figure 5.24). The thin, red-shaded area shows just
0
how much difference there is between these two integrals over the interval [0, 1].
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Chapter 5 | Integration
541
Figure 5.24 (a) The graph shows that over the interval
[0, 1], g(x) ≥ f (x), where equality holds only at the endpoints of the
interval. (b) Viewing the same graph with a greater zoom shows this more
clearly.
Average Value of a Function
We often need to find the average of a set of numbers, such as an average test grade. Suppose you received the following
test scores in your algebra class: 89, 90, 56, 78, 100, and 69. Your semester grade is your average of test scores and you
want to know what grade to expect. We can find the average by adding all the scores and dividing by the number of scores.
In this case, there are six test scores. Thus,
89 + 90 + 56 + 78 + 100 + 69 = 482 ≈ 80.33.
6
6
Therefore, your average test grade is approximately 80.33, which translates to a B− at most schools.
Suppose, however, that we have a function v(t) that gives us the speed of an object at any time t, and we want to find the
object’s average speed. The function v(t) takes on an infinite number of values, so we can’t use the process just described.
Fortunately, we can use a definite integral to find the average value of a function such as this.
Let f (x) be continuous over the interval ⎡⎣a, b⎤⎦ and let ⎡⎣a, b⎤⎦ be divided into n subintervals of width Δx = (b − a)/n.
Choose a representative x*i in each subinterval and calculate f ⎛⎝x*i
f ⎛⎝x*i
⎞
⎠
⎞
⎠
for i = 1, 2,…, n. In other words, consider each
as a sampling of the function over each subinterval. The average value of the function may then be approximated as
f ⎛⎝x*1 ⎞⎠ + f ⎛⎝x*2 ⎞⎠ + ⋯ + f ⎛⎝x*n ⎞⎠
,
n
which is basically the same expression used to calculate the average of discrete values.
a
b−a
But we know Δx = b −
n , so n = Δx , and we get
f ⎛⎝x*1 ⎞⎠ + f ⎛⎝x*2 ⎞⎠ + ⋯ + f ⎛⎝x*n ⎞⎠ f ⎛⎝x*1 ⎞⎠ + f ⎛⎝x*2 ⎞⎠ + ⋯ + f ⎛⎝x*n ⎞⎠
=
.
n
(b − a)
Δx
n
Following through with the algebra, the numerator is a sum that is represented as
∑
i=1
f ⎛⎝x*i ⎞⎠, and we are dividing by a
fraction. To divide by a fraction, invert the denominator and multiply. Thus, an approximate value for the average value of
the function is given by
542
Chapter 5 | Integration
n
∑
f ⎛⎝x*i
i=1
(b − a)
Δx
⎞
⎠
n
⎛
⎞
= ⎝ Δx ⎠ ∑ f ⎛⎝x*i
b−a i=1
⎞
⎠
n
⎞
⎛
= ⎝ 1 ⎠ ∑ f ⎛⎝x*i ⎞⎠Δx.
b−a i=1
This is a Riemann sum. Then, to get the exact average value, take the limit as n goes to infinity. Thus, the average value of
a function is given by
n
b
1 lim ∑ f (x )Δx = 1 ∫ f (x)dx.
i
b − a n → ∞i = 1
b−a a
Definition
Let f (x) be continuous over the interval ⎡⎣a, b⎤⎦. Then, the average value of the function f (x) (or fave) on ⎡⎣a, b⎤⎦ is
given by
f ave =
b
1
f (x)dx.
b − a∫a
Example 5.14
Finding the Average Value of a Linear Function
Find the average value of f (x) = x + 1 over the interval ⎡⎣0, 5⎤⎦.
Solution
First, graph the function on the stated interval, as shown in Figure 5.25.
Figure 5.25 The graph shows the area under the function
f (x) = x + 1 over ⎡⎣0, 5⎤⎦.
The region is a trapezoid lying on its side, so we can use the area formula for a trapezoid A = 1 h(a + b), where
2
h represents height, and a and b represent the two parallel sides. Then,
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Chapter 5 | Integration
543
∫
5
0
x + 1dx = 1 h(a + b)
2
1
= · 5 · (1 + 6)
2
= 35 .
2
Thus the average value of the function is
5
1
x + 1dx = 1 · 35 = 7 .
2
5 2
5 − 0∫0
5.13
Find the average value of f (x) = 6 − 2x over the interval [0, 3].
544
Chapter 5 | Integration
5.2 EXERCISES
In the following exercises, express the limits as integrals.
71.
n
∑ ⎛⎝x*i ⎞⎠Δx over [1, 3]
60. n lim
→∞
i=1
n
61. n lim
∑ ⎛⎝5⎛⎝x*i ⎞⎠2 − 3⎛⎝x*i ⎞⎠3⎞⎠Δx over [0, 2]
→∞
i=1
n
62. n lim
∑ sin 2 ⎛⎝2πx*i ⎞⎠Δx over [0, 1]
→∞
i=1
n
63. n lim
∑ cos 2 ⎛⎝2πx*i ⎞⎠Δx over [0, 1]
→∞
72.
i=1
In the following exercises, given Ln or Rn as indicated,
express their limits as n → ∞ as definite integrals,
identifying the correct intervals.
n
i−1
64. L n = 1
n∑ n
i=1
n
i
65. R n = 1
n∑ n
i=1
n
⎛
i − 1⎞
66. L n = 2
n ∑ ⎝1 + 2 n ⎠
i=1
n
⎛
i⎞
67. R n = 3
n ∑ ⎝3 + 3 n ⎠
i=1
n
i − 1 ⎛ i − 1⎞
68. L n = 2π
n ∑ 2π n cos⎝2π n ⎠
i=1
2
⎛
i ⎞ ⎛⎛
i⎞ ⎞
69. R n = 1
n ∑ ⎝1 + n ⎠log⎝⎝1 + n ⎠ ⎠
n
i=1
In the following exercises, evaluate the integrals of the
functions graphed using the formulas for areas of triangles
and circles, and subtracting the areas below the x-axis.
70.
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73.
Chapter 5 | Integration
74.
545
∫
83.
3
(3 − |x|)dx
−2
In the following exercises, use averages of values at the left
(L) and right (R) endpoints to compute the integrals of the
piecewise linear functions with graphs that pass through the
given list of points over the indicated intervals.
{(0, 0), (2, 1), (4, 3), (5, 0), (6, 0), (8, 3)} over
84.
[0, 8]
{(0, 2), (1, 0), (3, 5), (5, 5), (6, 2), (8, 0)} over
85.
[0, 8]
{(−4, −4), (−2, 0), (0, −2), (3, 3), (4, 3)} over
86.
75.
[−4, 4]
{(−4, 0), (−2, 2), (0, 0), (1, 2), (3, 2), (4, 0)}
87.
over [−4, 4]
Suppose that
∫
∫
4
0
f (x)dx = 5 and
4
0
g(x)dx = −1 and
∫
∫
2
0
f (x)dx = −3, and
2
0
g(x)dx = 2. In the following
exercises, compute the integrals.
In the following exercises, evaluate the integral using area
formulas.
76.
∫
77.
∫
78.
∫
79.
∫
80.
∫
81.
∫
82.
∫
3
0
(3 − x)dx
3
2
(3 − x)dx
3
(3 − |x|)dx
−3
6
0
(3 − |x − 3|)dx
2
−2
5
1
12
0
4 − x 2dx
4 − (x − 3) 2dx
36 − (x − 6) 2dx
88.
∫
89.
∫
90.
∫
91.
∫
92.
∫
93.
∫
In
∫
A
−A
4
⎛
⎝
0
4
⎛
⎝
2
2
⎛
⎝
0
4
⎛
⎝
2
2
f (x) + g(x)⎞⎠dx
f (x) − g(x)⎞⎠dx
f (x) − g(x)⎞⎠dx
⎛
⎝
3 f (x) − 4g(x)⎞⎠dx
0
4
2
f (x) + g(x)⎞⎠dx
⎛
⎝
the
4 f (x) − 3g(x)⎞⎠dx
following
f (x)dx = ∫
0
−A
exercises,
A
use
the
f (x)dx + ∫ f (x)dx to compute the
0
integrals.
π
94. ⌠
identity
sint dt (Hint: sin(−t) = −sin(t))
⌡−π 1 + t 2
546
Chapter 5 | Integration
95.
∫
96.
∫
97.
∫
In
∫
1
π
t
dt
π 1 + cost
−
109. Show that
3
4
(x − 3) 3 dx (Hint: Look at the graph of f.)
2
the
following
xdx = 1 ,
2
0
∫
exercises,
1
x 2 dx = 1 ,
3
0
given
∫
and
1
99.
100.
∫
∫
1
⎛
⎝1
0
1
0
∫
⎛
⎝1
x 3 dx = 1 ,
4
0
+ x + x 2 + x 3⎞⎠dx
− x + x − x ⎠dx
(1 − x) 2 dx
0
102.
⌠ ⎛6x −
⌡0⎝
1
∫
1
0
⎛
⎝7
105. Show that
∫
106. Show that
∫
107. Show that
108. Show that
⎡ π⎤
⎣0, 2 ⎦)
∫
3
⎛ 2
⎝x
− 6x + 9⎞⎠dx ≥ 0.
(x − 3)(x + 2)dx ≤ 0.
−2
1
0
2
1
∫
1 + x 3dx ≤ ∫
1 + xdx ≤ ∫
1
0
2
1
1 + x 2dx.
114.
f (x) = sin x, a = 0, b = 2π
115.
f (x) = cos x, a = 0, b = 2π
ln(256)
− 1.
3
2ln(2)
π .
119. [T] y = x + 1
4 − x2
exact solution is π .
6
over the interval [−1, 1]; the
In the following exercises, compute the average value using
the left Riemann sums LN for N = 1, 10, 100. How does
the accuracy compare with the given exact value?
120. [T] y = x 2 − 4 over the interval [0, 2]; the exact
solution is − 8 .
3
1 + x 2dx.
π/2
0
f (x) = (3 − |x|), a = −3, b = 3
solution is
In the following exercises, use the comparison
theorem.
3
113.
⎡
⎤
118. [T] y = tan x over the interval ⎣0, π ⎦; the exact
4
− 5x ⎠dx
0
f (x) = 4 − x 2, a = 0, b = 2
solution is 2( e − 1).
3⎞
∫
112.
117. [T] y = e x/2 over the interval [0, 1]; the exact
4 x 2⎞dx
3 ⎠
104. Show that
f (x) = x 5, a = −1, b = 1
solution is
(1 − 2x) 3 dx
0
111.
116. [T] y = ln(x) over the interval [1, 4]; the exact
1
∫
f (x) = x 2, a = −1, b = 1
In the following exercises, approximate the average value
using Riemann sums L100 and R100. How does your answer
compare with the exact given answer?
1
101.
103.
3⎞
2
costdt ≥ π 2/4.
−π/4
110.
that
compute the integrals.
98.
π/4
In the following exercises, find the average value fave of f
between a and b, and find a point c, where f (c) = f ave .
(2 − x)dx (Hint: Look at the graph of f.)
1
∫
sintdt ≥ π . (Hint: sint ≥ 2t
π over
4
This OpenStax book is available for free at http://cnx.org/content/col11964/1.2
121. [T] y = xe x
2
solution is 1 ⎛⎝e 4 − 1⎞⎠.
4
over the interval [0, 2]; the exact
Chapter 5 | Integration
547
⎛ ⎞
122. [T] y = ⎝1 ⎠
x
2
over the interval [0, 4]; the exact
15 .
64ln(2)
solution is
124.
B=∫
125.
0
A=∫
that
Suppose
A=∫
that
ai
ai − 1
such
that
either
f (t)dt.
a. Explain why
2π
0
sin 2 tdt
π/4
sec 2 tdt = π
−π/4
and
b
∫ a f (t)dt = A 1 + A 2 + ⋯ + A N .
b. Then, explain why
and
cos 2 tdt. Show that A + B = 2π and A = B.
π/4
Ai = ∫
cos⎛⎝π 2⎞⎠ − 1
.
2π
Suppose
2π
a = a0 < a1 < a2 < ⋯ < aN = b
f ≥ 0 over [a i − 1, a i] or f ≤ 0 over [a i − 1, a i]. Set
123. [T] y = x sin⎛⎝x 2⎞⎠ over the interval [−π, 0]; the
exact solution is
130. Suppose ⎡⎣a, b⎤⎦ can be subdivided into subintervals
|
b
∫ a f (t)dt
|
≤∫
b
a
| f (t)|dt.
131. Suppose f and g are continuous functions such that
d
∫c
⎡
⎣
d
f (t)dt ≤ ∫ g(t)dt for every subinterval ⎡⎣c, d⎤⎦ of
c
⎤
⎦
a, b . Explain why f (x) ≤ g(x) for all values of x.
132. Suppose the average value of f over ⎡⎣a, b⎤⎦ is 1 and
tan tdt. Show that A − B = π .
B=∫
2
−π/4
2
the average value of f over ⎡⎣b, c⎤⎦ is 1 where a < c < b.
Show that the average value of f over [a, c] is also 1.
2
126. Show that the average value of sin t over [0, 2π]
is equal to 1/2 Without further calculation, determine
whether the average value of sin 2 t over [0, π] is also
equal to 1/2.
127. Show that the average value of cos 2 t over [0, 2π]
is equal to 1/2. Without further calculation, determine
133.
Suppose that
⎡
⎣
a, b⎤⎦ can be partitioned. taking
a = a 0 < a 1 < ⋯ < a N = b such that the average value
of f over each subinterval [a i − 1, a i] = 1 is equal to 1 for
each i = 1,…, N. Explain why the average value of f over
⎡
⎣
a, b⎤⎦ is also equal to 1.
whether the average value of cos 2 (t) over [0, π] is also
134. Suppose that for each i such that 1 ≤ i ≤ N one has
equal to 1/2.
∫
128. Explain why the graphs of a quadratic function
(parabola) p(x) and a linear function ℓ(x) can intersect
in at most two points. Suppose that p(a) = ℓ(a) and
p(b) = ℓ(b), and that
why
d
d
∫ c p(t) > ∫ c ℓ(t)dt
b
b
∫ a p(t)dt > ∫ a ℓ(t)dt.
Explain
whenever a ≤ c < d ≤ b.
129. Suppose that parabola p(x) = ax + bx + c opens
downward (a < 0) and has a vertex of y = −b > 0. For
possible?
i−1
f (t)dt = i. Show that
∫
N
0
f (t)dt =
N(N + 1)
.
2
135. Suppose that for each i such that 1 ≤ i ≤ N one
has
∫
N
0
∫
f (t)dt =
i
i−1
f (t)dt = i 2.
Show
that
N(N + 1)(2N + 1)
.
6
136. [T] Compute the left and right Riemann sums L10
2
which interval [A, B] is
i
∫
B
A
⎛ 2
⎝ax
2a
+ bx + c⎞⎠dx as large as
L 10 + R 10
for f (t) = t 2 over
2
1
–
t 2 dt = 0.33, to how many
and R10 and their average
[0, 1]. Given that
decimal places is
∫
0
L 10 + R 10
accurate?
2
548
Chapter 5 | Integration
137. [T] Compute the left and right Riemann sums, L10
L + R 10
and R10, and their average 10
for f (t) = ⎛⎝4 − t 2⎞⎠
2
2
–
over [1, 2]. Given that ∫ ⎛⎝4 − t 2⎞⎠dt = 1.66, to how
1
many decimal places is
138.
∫
5
1
∫
If
5
L 10 + R 10
accurate?
2
1 + t 4dt = 41.7133...,
1
what
is
∫
1
tdt using the left and right endpoint
0
sums, each with a single rectangle. How does the average
of these left and right endpoint sums compare with the
actual value
∫
1
0
140. Estimate
tdt ?
∫
1
0
tdt by comparison with the area of a
single rectangle with height equal to the value of t at the
midpoint t = 1 . How does this midpoint estimate compare
2
with the actual value
∫
1
tdt ?
0
141. From the graph of sin(2πx) shown:
a. Explain why
∫
1
0
sin(2πt)dt = 0.
b. Explain why, in general,
a+1
∫a
sin(2πt)dt = 0 for
any value of a.
This OpenStax book is available for free at http://cnx.org/content/col11964/1.2
If f is 1-periodic
integrable
∫
1
0
143.
1 + u 4du ?
139. Estimate
142.
over
[0, 1],
⎛
⎝
f (t + 1) = f (t)⎞⎠,
is
it
If f is 1-periodic and
∫
always
odd, and
true
that
f (t)dt = 0 ?
necessarily true that
1+a
∫a
1
0
f (t)dt = A,
f (t)dt = A for all A?
is it
Chapter 5 | Integration
549
5.3 | The Fundamental Theorem of Calculus
Learning Objectives
5.3.1
5.3.2
5.3.3
5.3.4
5.3.5
5.3.6
Describe the meaning of the Mean Value Theorem for Integrals.
State the meaning of the Fundamental Theorem of Calculus, Part 1.
Use the Fundamental Theorem of Calculus, Part 1, to evaluate derivatives of integrals.
State the meaning of the Fundamental Theorem of Calculus, Part 2.
Use the Fundamental Theorem of Calculus, Part 2, to evaluate definite integrals.
Explain the relationship between differentiation and integration.
In the previous two sections, we looked at the definite integral and its relationship to the area under the curve of a function.
Unfortunately, so far, the only tools we have available to calculate the value of a definite integral are geometric area
formulas and limits of Riemann sums, and both approaches are extremely cumbersome. In this section we look at some
more powerful and useful techniques for evaluating definite integrals.
These new techniques rely on the relationship between differentiation and integration. This relationship was discovered and
explored by both Sir Isaac Newton and Gottfried Wilhelm Leibniz (among others) during the late 1600s and early 1700s,
and it is codified in what we now call the Fundamental Theorem of Calculus, which has two parts that we examine in this
section. Its very name indicates how central this theorem is to the entire development of calculus.
Isaac Newton’s contributions to mathematics and physics changed the way we look at the world. The relationships
he discovered, codified as Newton’s laws and the law of universal gravitation, are still taught as foundational
material in physics today, and his calculus has spawned entire fields of mathematics. To learn more, read a brief
biography (http://www.openstaxcollege.org/l/20_newtonbio) of Newton with multimedia clips.
Before we get to this crucial theorem, however, let’s examine another important theorem, the Mean Value Theorem for
Integrals, which is needed to prove the Fundamental Theorem of Calculus.
The Mean Value Theorem for Integrals
The Mean Value Theorem for Integrals states that a continuous function on a closed interval takes on its average value at
the same point in that interval. The theorem guarantees that if f (x) is continuous, a point c exists in an interval ⎡⎣a, b⎤⎦ such
that the value of the function at c is equal to the average value of f (x) over ⎡⎣a, b⎤⎦. We state this theorem mathematically
with the help of the formula for the average value of a function that we presented at the end of the preceding section.
Theorem 5.3: The Mean Value Theorem for Integrals
If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, then there is at least one point c ∈ ⎡⎣a, b⎤⎦ such that
f (c) =
b
1
f (x)dx.
b − a∫a
(5.15)
This formula can also be stated as
b
∫ a f (x)dx =
f (c)(b − a).
Proof
Since f (x) is continuous on ⎡⎣a, b⎤⎦, by the extreme value theorem (see Maxima and Minima), it assumes minimum and
maximum values—m and M, respectively—on ⎡⎣a, b⎤⎦. Then, for all x in ⎡⎣a, b⎤⎦, we have m ≤ f (x) ≤ M. Therefore, by
the comparison theorem (see The Definite Integral), we have
b
m(b − a) ≤ ∫ f (x)dx ≤ M(b − a).
a
550
Chapter 5 | Integration
Dividing by b − a gives us
m≤
Since
b
1
f (x)dx ≤ M.
b − a∫a
b
1
f (x)dx is a number between m and M, and since f (x) is continuous and assumes the values m and M over
b − a∫a
⎡
⎣
a, b⎤⎦, by the Intermediate Value Theorem (see Continuity), there is a number c over ⎡⎣a, b⎤⎦ such that
f (c) =
b
1
f (x)dx,
b − a∫a
and the proof is complete.
□
Example 5.15
Finding the Average Value of a Function
Find the average value of the function f (x) = 8 − 2x over the interval [0, 4] and find c such that f (c) equals
the average value of the function over [0, 4].
Solution
The formula states the mean value of f (x) is given by
4
1
(8 − 2x)dx.
4 − 0∫0
We can see in Figure 5.26 that the function represents a straight line and forms a right triangle bounded by the
x- and y-axes. The area of the triangle is A = 1 (base)⎛⎝height⎞⎠. We have
2
A = 1 (4)(8) = 16.
2
The average value is found by multiplying the area by 1/(4 − 0). Thus, the average value of the function is
1 (16) = 4.
4
Set the average value equal to f (c) and solve for c.
8 − 2c = 4
c = 2
At c = 2, f (2) = 4.
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Chapter 5 | Integration
551
Figure 5.26 By the Mean Value Theorem, the continuous
function f (x) takes on its average value at c at least once over
a closed interval.
Find the average value of the function f (x) = x over the interval ⎡⎣0, 6⎤⎦ and find c such that f (c)
2
equals the average value of the function over [0, 6].
5.14
Example 5.16
Finding the Point Where a Function Takes on Its Average Value
Given
∫
3
0
x 2 dx = 9, find c such that f (c) equals the average value of f (x) = x 2 over [0, 3].
Solution
We are looking for the value of c such that
f (c) =
3
1
x 2 dx = 1 (9) = 3.
3
3 − 0∫0
Replacing f (c) with c2, we have
c2 = 3
c = ± 3.
Since − 3 is outside the interval, take only the positive value. Thus, c = 3 (Figure 5.27).
552
Chapter 5 | Integration
Figure 5.27 Over the interval [0, 3], the function
f (x) = x 2 takes on its average value at c = 3.
5.15
Given
∫
3
0
⎛ 2
⎝2x
− 1⎞⎠dx = 15, find c such that f (c) equals the average value of f (x) = 2x 2 − 1 over
[0, 3].
Fundamental Theorem of Calculus Part 1: Integrals and
Antiderivatives
As mentioned earlier, the Fundamental Theorem of Calculus is an extremely powerful theorem that establishes the
relationship between differentiation and integration, and gives us a way to evaluate definite integrals without using Riemann
sums or calculating areas. The theorem is comprised of two parts, the first of which, the Fundamental Theorem of
Calculus, Part 1, is stated here. Part 1 establishes the relationship between differentiation and integration.
Theorem 5.4: Fundamental Theorem of Calculus, Part 1
If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, and the function F(x) is defined by
x
F(x) = ∫ f (t)dt,
(5.16)
a
then F′ (x) = f (x) over ⎡⎣a, b⎤⎦.
Before we delve into the proof, a couple of subtleties are worth mentioning here. First, a comment on the notation. Note that
we have defined a function, F(x), as the definite integral of another function, f (t), from the point a to the point x. At
first glance, this is confusing, because we have said several times that a definite integral is a number, and here it looks like
it’s a function. The key here is to notice that for any particular value of x, the definite integral is a number. So the function
F(x) returns a number (the value of the definite integral) for each value of x.
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Chapter 5 | Integration
553
Second, it is worth commenting on some of the key implications of this theorem. There is a reason it is called the
Fundamental Theorem of Calculus. Not only does it establish a relationship between integration and differentiation, but
also it guarantees that any integrable function has an antiderivative. Specifically, it guarantees that any continuous function
has an antiderivative.
Proof
Applying the definition of the derivative, we have
F′ (x) = lim
h→0
F(x + h) − F(x)
h
⎡
x+h
= lim 1 ⎢∫
h → 0 h⎣ a
⎡ x+h
= lim 1 ⎢
f (t)dt +
h → 0 h⎣ a
∫
x+h
= lim 1 ∫
h→0h x
Looking carefully at this last expression, we see 1 ∫
h
x+h
x
x
⎤
a
⎦
f (t)dt − ∫ f (t)dt⎥
a
∫x
⎤
f (t)dt⎥
⎦
f (t)dt.
f (t)dt is just the average value of the function f (x) over the
interval ⎡⎣x, x + h⎤⎦. Therefore, by The Mean Value Theorem for Integrals, there is some number c in ⎡⎣x, x + h⎤⎦ such
that
x+h
1∫
h x
f (x)dx = f (c).
In addition, since c is between x and h, c approaches x as h approaches zero. Also, since f (x) is continuous, we have
lim f (c) = clim
f (c) = f (x). Putting all these pieces together, we have
→x
h→0
x+h
F′ (x) = lim 1 ∫
h→0h x
= lim f (c)
f (x)dx
h→0
= f (x),
and the proof is complete.
□
Example 5.17
Finding a Derivative with the Fundamental Theorem of Calculus
Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of
x
g(x) = ⌠ 3 1 dt.
⌡1 t + 1
Solution
According to the Fundamental Theorem of Calculus, the derivative is given by
g′ (x) =
1 .
x3 + 1
554
Chapter 5 | Integration
5.16
Use the Fundamental Theorem of Calculus, Part 1 to find the derivative of g(r) = ∫
r
0
x 2 + 4dx.
Example 5.18
Using the Fundamental Theorem and the Chain Rule to Calculate Derivatives
Let F(x) = ∫
x
1
sintdt. Find F′ (x).
Solution
Letting u(x) = x, we have F(x) = ∫
u(x)
1
sintdt. Thus, by the Fundamental Theorem of Calculus and the chain
rule,
F′ (x) = sin⎛⎝u(x)⎞⎠ du
dx
⎞
⎛
= sin(u(x)) · ⎝1 x −1/2⎠
2
sin
x
.
=
2 x
5.17
Let F(x) = ∫
x3
1
costdt. Find F′ (x).
Example 5.19
Using the Fundamental Theorem of Calculus with Two Variable Limits of
Integration
Let F(x) = ∫
2x
x
t 3 dt. Find F′ (x).
Solution
We have F(x) = ∫
2x
x
t 3 dt. Both limits of integration are variable, so we need to split this into two integrals. We
get
F(x) = ∫
2x
x
0
t 3 dt
= ∫ t 3 dt + ∫
x
x
2x
0
= −∫ t 3 dt + ∫
0
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t 3 dt
2x
0
t 3 dt.
Chapter 5 | Integration
555
Differentiating the first term, we obtain
d ⎡− t 3 dt⎤ = −x 3.
⎦
dx ⎣ ∫ 0
x
Differentiating the second term, we first let u(x) = 2x. Then,
⎡
2x
⎤
⎡ u(x)
⎤
t 3 dt⎥
d⎢
t 3 dt⎥ = d ⎢∫
dx ⎣ 0
dx ⎣ ∫ 0
⎦
⎦
= (u(x)) 3 du
dx
3
= (2x) · 2
= 16x 3.
Thus,
⎤
⎡
F′ (x) = d −∫ t 3 dt + d ⎢ ∫ t 3 dt⎥
⎦ dx ⎣ 0
⎣
dx
⎦
0
x
⎡
2x
⎤
= −x 3 + 16x 3
= 15x 3.
5.18
Let F(x) = ∫
x2
x
costdt. Find F′ (x).
Fundamental Theorem of Calculus, Part 2: The Evaluation Theorem
The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. After tireless efforts
by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools
to explain many phenomena. Using calculus, astronomers could finally determine distances in space and map planetary
orbits. Everyday financial problems such as calculating marginal costs or predicting total profit could now be handled with
simplicity and accuracy. Engineers could calculate the bending strength of materials or the three-dimensional motion of
objects. Our view of the world was forever changed with calculus.
After finding approximate areas by adding the areas of n rectangles, the application of this theorem is straightforward by
comparison. It almost seems too simple that the area of an entire curved region can be calculated by just evaluating an
antiderivative at the first and last endpoints of an interval.
Theorem 5.5: The Fundamental Theorem of Calculus, Part 2
If f is continuous over the interval ⎡⎣a, b⎤⎦ and F(x) is any antiderivative of f (x), then
b
∫ a f (x)dx = F(b) − F(a).
(5.17)
We often see the notation F(x)| ba to denote the expression F(b) − F(a). We use this vertical bar and associated limits a
and b to indicate that we should evaluate the function F(x) at the upper limit (in this case, b), and subtract the value of the
function F(x) evaluated at the lower limit (in this case, a).
The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an
556
Chapter 5 | Integration
antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints
of the interval and subtracting.
Proof
Let P = {x i}, i = 0, 1,…, n be a regular partition of ⎡⎣a, b⎤⎦. Then, we can write
F(b) − F(a) = F(x n) − F(x 0)
= ⎡⎣F(x n) − F(x n − 1)⎤⎦ + ⎡⎣F(x n − 1) − F(x n − 2)⎤⎦ + … + ⎡⎣F(x 1) − F(x 0)⎤⎦
=
n
∑
i=1
⎡
⎣
F(x i) − F(x i − 1)⎤⎦.
Now, we know F is an antiderivative of f over ⎡⎣a, b⎤⎦, so by the Mean Value Theorem (see The Mean Value Theorem)
for i = 0, 1,…, n we can find c i in [x i − 1, x i] such that
F(x i) − F(x i − 1) = F′ (c i⎞⎠(x i − x i − 1) = f (c i)Δx.
Then, substituting into the previous equation, we have
F(b) − F(a) =
n
∑
i=1
f (c i)Δx.
Taking the limit of both sides as n → ∞, we obtain
n
F(b) − F(a) = n lim
∑ f (c i)Δx
→∞
b
i=1
= ∫ f (x)dx.
a
□
Example 5.20
Evaluating an Integral with the Fundamental Theorem of Calculus
Use The Fundamental Theorem of Calculus, Part 2 to evaluate
∫
2
⎛ 2
⎝t
−2
− 4⎞⎠dt.
Solution
Recall the power rule for Antiderivatives:
If y = x n,
n+1
∫ x n dx = xn + 1 + C.
Use this rule to find the antiderivative of the function and then apply the theorem. We have
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Chapter 5 | Integration
557
∫
2
⎛ 2
⎝t
−2
3
− 4⎞⎠dt = t − 4t| 2−2
3
=
⎡(2) 3
⎤ ⎡(−2) 3
⎤
− 4(2) −
⎣ 3
⎦ ⎣ 3 − 4(−2)⎦
⎛
⎞ ⎛
⎞
= ⎝8 − 8⎠ − ⎝− 8 + 8⎠
3
3
= 8 −8+ 8 −8
3
3
16
=
− 16
3
= − 32 .
3
Analysis
Notice that we did not include the “+ C” term when we wrote the antiderivative. The reason is that, according
to the Fundamental Theorem of Calculus, Part 2, any antiderivative works. So, for convenience, we chose the
antiderivative with C = 0. If we had chosen another antiderivative, the constant term would have canceled out.
This always happens when evaluating a definite integral.
The region of the area we just calculated is depicted in Figure 5.28. Note that the region between the curve
and the x-axis is all below the x-axis. Area is always positive, but a definite integral can still produce a negative
number (a net signed area). For example, if this were a profit function, a negative number indicates the company
is operating at a loss over the given interval.
Figure 5.28 The evaluation of a definite integral can produce
a negative value, even though area is always positive.
Example 5.21
Evaluating a Definite Integral Using the Fundamental Theorem of Calculus, Part 2
Evaluate the following integral using the Fundamental Theorem of Calculus, Part 2:
∫
Solution
9
1
x − 1 dx.
x
558
Chapter 5 | Integration
First, eliminate the radical by rewriting the integral using rational exponents. Then, separate the numerator terms
by writing each one over the denominator:
⌠ x − 1 dx = ⌠ ⎛ x − 1 ⎞dx.
⌡1 x 1/2
⌡1⎝x 1/2 x 1/2 ⎠
9
9
Use the properties of exponents to simplify:
⌠ ⎛ x − 1 ⎞dx = ∫ ⎛x 1/2 − x −1/2⎞dx.
⎠
⎝
⌡1⎝x 1/2 x 1/2 ⎠
1
9
9
Now, integrate using the power rule:
∫
9
1
⎛ 1/2
⎝x
−x
|
⎛ 3/2
⎞
x
x 1/2
⎠dx = ⎜ 3 − 1 ⎟ 1
⎝ 2
2 ⎠
−1/2⎞
9
⎡ 3/2
⎤ ⎡ 3/2
⎤
(1)
(9)
(9) 1/2
(1) 1/2
= ⎢ 3 − 1 ⎥− ⎢ 3 − 1 ⎥
⎣ 2
2 ⎦ ⎣ 2
2 ⎦
⎡
⎤ ⎡
⎤
= ⎣2 (27) − 2(3)⎦ − ⎣2 (1) − 2(1)⎦
3
3
= 18 − 6 − 2 + 2
3
40
= .
3
See Figure 5.29.
Figure 5.29 The area under the curve from x = 1 to x = 9
can be calculated by evaluating a definite integral.
5.19
Use The Fundamental Theorem of Calculus, Part 2 to evaluate
∫
2
1
x −4 dx.
Example 5.22
A Roller-Skating Race
James and Kathy are racing on roller skates. They race along a long, straight track, and whoever has gone the
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Chapter 5 | Integration
559
farthest after 5 sec wins a prize. If James can skate at a velocity of f (t) = 5 + 2t ft/sec and Kathy can skate at a
⎛
⎞
velocity of g(t) = 10 + cos⎝π t⎠ ft/sec, who is going to win the race?
2
Solution
We need to integrate both functions over the interval ⎡⎣0, 5⎤⎦ and see which value is bigger. For James, we want to
calculate
∫
5
0
(5 + 2t)dt.
Using the power rule, we have
∫
5
0
|
5
(5 + 2t)dt = ⎛⎝5t + t 2⎞⎠ 0
= (25 + 25) = 50.
Thus, James has skated 50 ft after 5 sec. Turning now to Kathy, we want to calculate
∫
⎛
5
⎞
10 + cos⎝π t⎠dt.
2
0
⎛
⎞
We know sint is an antiderivative of cost, so it is reasonable to expect that an antiderivative of cos⎝π t⎠ would
2
⎛ ⎞
⎛ ⎞
⎛ ⎞
involve sin⎝π t⎠. However, when we differentiate sin⎝π t⎠, we get π cos⎝π t⎠ as a result of the chain rule, so we
2
2
2
2
have to account for this additional coefficient when we integrate. We obtain
∫
|
⎛ ⎞
⎛ ⎞⎞
⎛
10 + cos⎝π t⎠dt = ⎝10t + π2 sin⎝π t⎠⎠ 0
2
2
0
⎛
⎞ ⎛
2
= ⎝50 + π ⎠ − ⎝0 − π2 sin0⎞⎠
≈ 50.6.
5
5
Kathy has skated approximately 50.6 ft after 5 sec. Kathy wins, but not by much!
5.20 Suppose James and Kathy have a rematch, but this time the official stops the contest after only 3 sec.
Does this change the outcome?
560
Chapter 5 | Integration
A Parachutist in Free Fall
Figure 5.30 Skydivers can adjust the velocity of their dive by changing the position of their body during the
free fall. (credit: Jeremy T. Lock)
Julie is an avid skydiver. She has more than 300 jumps under her belt and has mastered the art of making adjustments
to her body position in the air to control how fast she falls. If she arches her back and points her belly toward the
ground, she reaches a terminal velocity of approximately 120 mph (176 ft/sec). If, instead, she orients her body with
her head straight down, she falls faster, reaching a terminal velocity of 150 mph (220 ft/sec).
Since Julie will be moving (falling) in a downward direction, we assume the downward direction is positive to simplify
our calculations. Julie executes her jumps from an altitude of 12,500 ft. After she exits the aircraft, she immediately
starts falling at a velocity given by v(t) = 32t. She continues to accelerate according to this velocity function until she
reaches terminal velocity. After she reaches terminal velocity, her speed remains constant until she pulls her ripcord
and slows down to land.
On her first jump of the day, Julie orients herself in the slower “belly down” position (terminal velocity is 176 ft/sec).
Using this information, answer the following questions.
1. How long after she exits the aircraft does Julie reach terminal velocity?
2. Based on your answer to question 1, set up an expression involving one or more integrals that represents the
distance Julie falls after 30 sec.
3. If Julie pulls her ripcord at an altitude of 3000 ft, how long does she spend in a free fall?
4. Julie pulls her ripcord at 3000 ft. It takes 5 sec for her parachute to open completely and for her to slow down,
during which time she falls another 400 ft. After her canopy is fully open, her speed is reduced to 16 ft/sec.
Find the total time Julie spends in the air, from the time she leaves the airplane until the time her feet touch the
ground.
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Chapter 5 | Integration
561
On Julie’s second jump of the day, she decides she wants to fall a little faster and orients herself in the “head
down” position. Her terminal velocity in this position is 220 ft/sec. Answer these questions based on this
velocity:
5. How long does it take Julie to reach terminal velocity in this case?
6. Before pulling her ripcord, Julie reorients her body in the “belly down” position so she is not moving quite as
fast when her parachute opens. If she begins this maneuver at an altitude of 4000 ft, how long does she spend
in a free fall before beginning the reorientation?
Some jumpers wear “ wingsuits” (see Figure 5.31). These suits have fabric panels between the arms and legs
and allow the wearer to glide around in a free fall, much like a flying squirrel. (Indeed, the suits are sometimes
called “flying squirrel suits.”) When wearing these suits, terminal velocity can be reduced to about 30 mph (44
ft/sec), allowing the wearers a much longer time in the air. Wingsuit flyers still use parachutes to land; although
the vertical velocities are within the margin of safety, horizontal velocities can exceed 70 mph, much too fast
to land safely.
Figure 5.31 The fabric panels on the arms and legs of a wingsuit work to reduce the vertical velocity of a
skydiver’s fall. (credit: Richard Schneider)
Answer the following question based on the velocity in a wingsuit.
7. If Julie dons a wingsuit before her third jump of the day, and she pulls her ripcord at an altitude of 3000 ft, how
long does she get to spend gliding around in the air?
562
Chapter 5 | Integration
5.3 EXERCISES
144.
Consider two athletes running at variable speeds
v 1 (t) and v 2 (t). The runners start and finish a race at
d⌠
t dt
dx ⌡1 1 + t
158.
d
e t dt
dx ∫ 0
159.
d
lnu 2 du
dx ∫ 1
exactly the same time. Explain why the two runners must
be going the same speed at some point.
145. Two mountain climbers start their climb at base
camp, taking two different routes, one steeper than the
other, and arrive at the peak at exactly the same time. Is it
necessarily true that, at some point, both climbers increased
in altitude at the same rate?
146. To get on a certain toll road a driver has to take a
card that lists the mile entrance point. The card also has a
timestamp. When going to pay the toll at the exit, the driver
is surprised to receive a speeding ticket along with the toll.
Explain how this can happen.
147.
x2
157.
ln x
e2
x
160. The graph of y = ∫ f (t)dt, where f is a piecewise
0
constant function, is shown here.
x
Set F(x) = ∫ (1 − t)dt. Find F′ (2) and the
1
′
average value of F over [1, 2].
In the following exercises, use the Fundamental Theorem
of Calculus, Part 1, to find each derivative.
x
−t 2
148.
d
e
dx ∫ 1
149.
d
e cost dt
dx ∫ 1
dt
x
x
150. d ∫ 9 − y 2dy
dx 3
a. Over which intervals is f positive? Over which
intervals is it negative? Over which intervals, if
any, is it equal to zero?
b. What are the maximum and minimum values of f?
c. What is the average value of f?
x
161. The graph of y = ∫ f (t)dt, where f is a piecewise
0
constant function, is shown here.
x
151.
d⌠
ds
dx ⌡ 16 − s 2
4
152.
d
tdt
dx ∫ x
153.
d
tdt
dx ∫ 0
154.
d
dx ∫ 0
155.
d
1 − t 2dt
dx ∫ cos x
156.
d⌠
t 2 dt
dx ⌡ 1 + t 4
1
2x
x
sin x
1 − t 2dt
1
x
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a. Over which intervals is f positive? Over which
intervals is it negative? Over which intervals, if
any, is it equal to zero?
b. What are the maximum and minimum values of f?
c. What is the average value of f?
Chapter 5 | Integration
563
x
162. The graph of y = ∫ ℓ(t)dt, where ℓ is a piecewise
0
linear function, is shown here.
169. [T] ⌠ 42 dx over [1, 4]
⌡x
In the following exercises, evaluate each definite integral
using the Fundamental Theorem of Calculus, Part 2.
a. Over which intervals is ℓ positive? Over which
intervals is it negative? Over which, if any, is it
zero?
b. Over which intervals is ℓ increasing? Over which is
it decreasing? Over which, if any, is it constant?
c. What is the average value of ℓ?
x
163. The graph of y = ∫ ℓ(t)dt, where ℓ is a piecewise
170.
∫
171.
∫
172.
∫
173.
∫
174.
∫
175.
∫
176.
∫
0
linear function, is shown here.
2
⎛ 2
⎝x
− 3x⎞⎠dx
⎛ 2
⎝x
+ 3x − 5⎞⎠dx
−1
3
−2
3
(t + 2)(t − 3)dt
−2
3
2
⎛ 2
⎝t
2
− 9⎞⎠⎛⎝4 − t 2⎞⎠dt
x 9 dx
1
1
0
x 99 dx
8
4
⎛ 5/2
⎝4t
− 3t 3/2⎞⎠dt
⎛ 2 1⎞
x − 2 dx
⌡1/4⎝
x ⎠
4
177. ⌠
a. Over which intervals is ℓ positive? Over which
intervals is it negative? Over which, if any, is it
zero?
b. Over which intervals is ℓ increasing? Over which
is it decreasing? Over which intervals, if any, is it
constant?
c. What is the average value of ℓ?
2
178. ⌠ 23 dx
⌡1 x
179.
4
⌠ 1 dx
⌡1 2 x
In the following exercises, use a calculator to estimate the
area under the curve by computing T10, the average of
the left- and right-endpoint Riemann sums using N = 10
rectangles. Then, using the Fundamental Theorem of
Calculus, Part 2, determine the exact area.
180. ⌠ 2 −2 t dt
⌡ t
164. [T] y = x 2 over [0, 4]
181. ⌠
4
1
16
dt
⌡1 t 1/4
165. [T] y = x 3 + 6x 2 + x − 5 over [−4, 2]
166. [T] y = x 3 over ⎡⎣0, 6⎤⎦
167. [T] y = x + x 2 over [1, 9]
168. [T]
∫ (cos x − sin x)dx
over [0, π]
182.
∫
183.
∫
2π
0
cosθdθ
π/2
0
sinθdθ
564
184.
185.
186.
187.
Chapter 5 | Integration
∫
∫
∫
π/4
198. Suppose that the number of hours of daylight on
a given day in Seattle is modeled by the function
sec 2 θdθ
0
⎛
π/4
secθ tanθ
0
a. What is the average number of daylight hours in a
year?
b. At
which
times
t1
and
t 2,
where
0 ≤ t 1 < t 2 < 12, do the number of daylight
π/4
cscθ cot θdθ
π/3
csc θdθ
hours equal the average number?
c. Write an integral that expresses the total number of
daylight hours in Seattle between t 1 and t 2.
⎛1 1 ⎞
dt
−
⌡ ⎝t 2 t 3 ⎠
d. Compute the mean hours of daylight in Seattle
between t 1 and t 2, where 0 ≤ t 1 < t 2 < 12,
and then between t 2 and t 1, and show that the
∫
π/2
π/4
2
2
188. ⌠
1
⎛1 1 ⎞
−
dt
⌡−2⎝t 2 t 3 ⎠
average of the two is equal to the average day
length.
−1
189. ⌠
In the following exercises, use the evaluation theorem to
express the integral as a function F(x).
199. Suppose the rate of gasoline consumption in the
United States can be modeled by a sinusoidal function of
⎛
∫ a t 2 dt
191.
∫
192.
∫
x
1
x
0
200. Explain why, if f is continuous over ⎡⎣a, b⎤⎦, there
∫ −xsintdt
is
In the following exercises, identify the roots of the
integrand to remove absolute values, then evaluate using
the Fundamental Theorem of Calculus, Part 2.
∫
195.
∫
196.
∫
197.
∫
|x|dx
|t
2
−2
π
0
−π/2
f (c) =
least
one
point
c ∈ ⎡⎣a, b⎤⎦
such
that
b
1
f (t)dt.
b − a∫a
201. Explain why, if f is continuous over ⎡⎣a, b⎤⎦ and is not
such that f (M) =
|
− 2t − 3 dt
|cost|dt
π/2
at
equal to a constant, there is at least one point M ∈ ⎡⎣a, b⎤⎦
−2
4
⎞⎞
and the end of September ⎛⎝t = 9).
costdt
x
194.
6
a. What is the average monthly consumption, and for
which values of t is the rate at time t equal to the
average rate?
b. What is the number of gallons of gasoline
consumed in the United States in a year?
c. Write an integral that expresses the average
monthly U.S. gas consumption during the part of
the year between the beginning of April (t = 3)
e t dt
3
⎛
the form ⎝11.21 − cos⎝πt ⎠⎠ × 10 9 gal/mo.
x
190.
193.
⎞
−3.75cos⎝πt ⎠ + 12.25, with t given in months and
6
t = 0 corresponding to the winter solstice.
|sint|dt
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b
1
f (t)dt and at least one point
b − a∫a
m ∈ ⎡⎣a, b⎤⎦ such that f (m) <
b
1
f (t)dt.
b − a∫a
Chapter 5 | Integration
565
202. Kepler’s first law states that the planets move in
elliptical orbits with the Sun at one focus. The closest point
of a planetary orbit to the Sun is called the perihelion (for
Earth, it currently occurs around January 3) and the farthest
point is called the aphelion (for Earth, it currently occurs
around July 4). Kepler’s second law states that planets
sweep out equal areas of their elliptical orbits in equal
times. Thus, the two arcs indicated in the following figure
are swept out in equal times. At what time of year is Earth
moving fastest in its orbit? When is it moving slowest?
203. A point on an ellipse with major axis length 2a
and minor axis length 2b has the coordinates
(acosθ, bsinθ), 0 ≤ θ ≤ 2π.
a. Show that the distance from this point to the focus
at (−c, 0) is d(θ) = a + ccosθ,
where
c = a 2 − b 2.
b. Use these coordinates to show that the average
–
distance d from a point on the ellipse to the focus
at (−c, 0), with respect to angle θ, is a.
204. As implied earlier, according to Kepler’s laws,
Earth’s orbit is an ellipse with the Sun at one focus. The
perihelion for Earth’s orbit around the Sun is 147,098,290
km and the aphelion is 152,098,232 km.
a. By placing the major axis along the x-axis, find the
average distance from Earth to the Sun.
b. The classic definition of an astronomical unit (AU)
is the distance from Earth to the Sun, and its value
was computed as the average of the perihelion and
aphelion distances. Is this definition justified?
205. The force of gravitational attraction between the Sun
, where m is the mass of the
and a planet is F(θ) = GmM
2
r (θ)
planet, M is the mass of the Sun, G is a universal constant,
and r(θ) is the distance between the Sun and the planet
when the planet is at an angle θ with the major axis of its
orbit. Assuming that M, m, and the ellipse parameters a
and b (half-lengths of the major and minor axes) are given,
set up—but do not evaluate—an integral that expresses in
terms of G, m, M, a, b the average gravitational force
between the Sun and the planet.
206. The displacement from rest of a mass attached to
a spring satisfies the simple harmonic motion equation
x(t) = Acos⎛⎝ωt − ϕ⎞⎠, where ϕ is a phase constant, ω is
the angular frequency, and A is the amplitude. Find the
average velocity, the average speed (magnitude of
velocity), the average displacement, and the average
distance from rest (magnitude of displacement) of the
mass.
566
Chapter 5 | Integration
5.4 | Integration Formulas and the Net Change Theorem
Learning Objectives
5.4.1
5.4.2
5.4.3
5.4.4
Apply the basic integration formulas.
Explain the significance of the net change theorem.
Use the net change theorem to solve applied problems.
Apply the integrals of odd and even functions.
In this section, we use some basic integration formulas studied previously to solve some key applied problems. It is
important to note that these formulas are presented in terms of indefinite integrals. Although definite and indefinite integrals
are closely related, there are some key differences to keep in mind. A definite integral is either a number (when the limits
of integration are constants) or a single function (when one or both of the limits of integration are variables). An indefinite
integral represents a family of functions, all of which differ by a constant. As you become more familiar with integration,
you will get a feel for when to use definite integrals and when to use indefinite integrals. You will naturally select the correct
approach for a given problem without thinking too much about it. However, until these concepts are cemented in your mind,
think carefully about whether you need a definite integral or an indefinite integral and make sure you are using the proper
notation based on your choice.
Basic Integration Formulas
Recall the integration formulas given in Table 4.27 and the rule on properties of definite integrals. Let’s look at a few
examples of how to apply these rules.
Example 5.23
Integrating a Function Using the Power Rule
Use the power rule to integrate the function
∫
4
1
t(1 + t)dt.
Solution
The first step is to rewrite the function and simplify it so we can apply the power rule:
∫
4
1
4
t(1 + t)dt = ∫ t 1/2(1 + t)dt
1
=∫
4
1
⎛ 1/2
⎝t
+ t 3/2⎞⎠dt.
Now apply the power rule:
∫
5.21
4
1
⎛ 1/2
⎝t
|
⎛
⎞
+ t 3/2⎞⎠dt = ⎝2 t 3/2 + 2 t 5/2⎠ 1
3
5
4
⎡
⎤ ⎡
⎤
= ⎣2 (4) 3/2 + 2 (4) 5/2⎦ − ⎣2 (1) 3/2 + 2 (1) 5/2⎦
3
3
5
5
= 256 .
15
Find the definite integral of f (x) = x 2 − 3x over the interval [1, 3].
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Chapter 5 | Integration
567
The Net Change Theorem
The net change theorem considers the integral of a rate of change. It says that when a quantity changes, the new value
equals the initial value plus the integral of the rate of change of that quantity. The formula can be expressed in two ways.
The second is more familiar; it is simply the definite integral.
Theorem 5.6: Net Change Theorem
The new value of a changing quantity equals the initial value plus the integral of the rate of change:
b
F(b) = F(a) + ∫ F '(x)dx
(5.18)
a
b
or
∫ a F '(x)dx = F(b) − F(a).
Subtracting F(a) from both sides of the first equation yields the second equation. Since they are equivalent formulas, which
one we use depends on the application.
The significance of the net change theorem lies in the results. Net change can be applied to area, distance, and volume, to
name only a few applications. Net change accounts for negative quantities automatically without having to write more than
one integral. To illustrate, let’s apply the net change theorem to a velocity function in which the result is displacement.
We looked at a simple example of this in The Definite Integral. Suppose a car is moving due north (the positive direction)
at 40 mph between 2 p.m. and 4 p.m., then the car moves south at 30 mph between 4 p.m. and 5 p.m. We can graph this
motion as shown in Figure 5.32.
Figure 5.32 The graph shows speed versus time for the given
motion of a car.
Just as we did before, we can use definite integrals to calculate the net displacement as well as the total distance traveled.
The net displacement is given by
∫
2
5
v(t)dt = ∫ 40dt + ⌠ −30dt
5
4
2
= 80 − 30
= 50.
⌡4
Thus, at 5 p.m. the car is 50 mi north of its starting position. The total distance traveled is given by
568
Chapter 5 | Integration
5
∫ |v(t)|dt = ∫ 40dt + ⌠ 30dt
2
2
⌡4
5
4
= 80 + 30
= 110.
Therefore, between 2 p.m. and 5 p.m., the car traveled a total of 110 mi.
To summarize, net displacement may include both positive and negative values. In other words, the velocity function
accounts for both forward distance and backward distance. To find net displacement, integrate the velocity function over
the interval. Total distance traveled, on the other hand, is always positive. To find the total distance traveled by an object,
regardless of direction, we need to integrate the absolute value of the velocity function.
Example 5.24
Finding Net Displacement
Given a velocity function v(t) = 3t − 5 (in meters per second) for a particle in motion from time t = 0 to time
t = 3, find the net displacement of the particle.
Solution
Applying the net change theorem, we have
∫
3
2
(3t − 5)dt = 3t − 5t
2
0
=
|
3
0
⎤
⎡3(3)
⎣ 2 − 5(3)⎦ − 0
2
= 27 − 15
2
= 27 − 30
2
2
3
= − .
2
The net displacement is − 3 m (Figure 5.33).
2
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Chapter 5 | Integration
569
Figure 5.33 The graph shows velocity versus time for a
particle moving with a linear velocity function.
Example 5.25
Finding the Total Distance Traveled
Use Example 5.24 to find the total distance traveled by a particle according to the velocity function
v(t) = 3t − 5 m/sec over a time interval [0, 3].
Solution
The total distance traveled includes both the positive and the negative values. Therefore, we must integrate the
absolute value of the velocity function to find the total distance traveled.
To continue with the example, use two integrals to find the total distance. First, find the t-intercept of the function,
since that is where the division of the interval occurs. Set the equation equal to zero and solve for t. Thus,
3t − 5 = 0
3t = 5
t = 5.
3
⎡
⎤
⎡
⎤
The two subintervals are ⎣0, 5 ⎦ and ⎣ 5 , 3⎦. To find the total distance traveled, integrate the absolute value of
3
3
⎡
⎤
the function. Since the function is negative over the interval ⎣0, 5 ⎦, we have |v(t)| = −v(t) over that interval.
3
⎡
⎤
Over ⎣ 5 , 3⎦, the function is positive, so |v(t)| = v(t). Thus, we have
3
570
Chapter 5 | Integration
5/3
∫ |v(t)|dt = ⌠ −v(t)dt + ∫ v(t)dt
0
5/3
⌡0
3
=∫
3
5/3
5 − 3tdt + ∫
0
|
3
3t − 5dt
5/3
|
2⎞
⎞
⎛
⎛ 2
= ⎝5t − 3t ⎠ 0 + ⎝3t − 5t⎠ 5/3
2
2
5/3
3
⎡ ⎛ ⎞ 3(5/3) 2 ⎤
⎡
⎤ ⎡3(5/3) 2 25 ⎤
= 5⎝5 ⎠ −
− 0 + ⎣27 − 15⎦ −
⎣ 3
⎣ 2 − 3⎦
2 ⎦
2
= 25 − 25 + 27 − 15 − 25 + 25
3
3
2
6
6
41
= .
6
So, the total distance traveled is 14 m.
6
5.22 Find the net displacement and total distance traveled in meters given the velocity function
f (t) = 1 e t − 2 over the interval [0, 2].
2
Applying the Net Change Theorem
The net change theorem can be applied to the flow and consumption of fluids, as shown in Example 5.26.
Example 5.26
How Many Gallons of Gasoline Are Consumed?
If the motor on a motorboat is started at t = 0 and the boat consumes gasoline at the rate of 5 − t 3 gal/hr, how
much gasoline is used in the first 2 hours?
Solution
Express the problem as a definite integral, integrate, and evaluate using the Fundamental Theorem of Calculus.
The limits of integration are the endpoints of the interval [0, 2]. We have
∫
2
0
⎛
⎝5
⎛
|
4⎞
− t 3⎞⎠dt = ⎝5t − t ⎠
4
⎡
⎣
= 5(2) −
= 10 − 16
4
= 6.
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2
0
(2) 4 ⎤
−0
4 ⎦
Chapter 5 | Integration
571
Thus, the motorboat uses 6 gal of gas in 2 hours.
Example 5.27
Chapter Opener: Iceboats
Figure 5.34 (credit: modification of work by Carter Brown,
Flickr)
As we saw at the beginning of the chapter, top iceboat racers (Figure 5.1) can attain speeds of up to five times the
wind speed. Andrew is an intermediate iceboater, though, so he attains speeds equal to only twice the wind speed.
Suppose Andrew takes his iceboat out one morning when a light 5-mph breeze has been blowing all morning. As
Andrew gets his iceboat set up, though, the wind begins to pick up. During his first half hour of iceboating, the
wind speed increases according to the function v(t) = 20t + 5. For the second half hour of Andrew’s outing, the
wind remains steady at 15 mph. In other words, the wind speed is given by
⎧20t + 5
v(t) = ⎨
⎩15
for 0 ≤ t ≤ 1
2
1
for ≤ t ≤ 1.
2
Recalling that Andrew’s iceboat travels at twice the wind speed, and assuming he moves in a straight line away
from his starting point, how far is Andrew from his starting point after 1 hour?
Solution
To figure out how far Andrew has traveled, we need to integrate his velocity, which is twice the wind speed. Then
1
Distance = ∫ 2v(t)dt.
0
Substituting the expressions we were given for v(t), we get
572
Chapter 5 | Integration
∫
1/2
2v(t)dt = ⌠ 2v(t)dt + ∫
1
0
⌡0
1
2v(t)dt
1/2
1/2
= ⌠ 2(20t + 5)dt + ∫
⌡0
1/2
1
2(15)dt
1/3
= ⌠ (40t + 10)dt + ∫ 30dt
1/2
⌡0
1
+ [30t]| 11/2
= ⎡⎣20t 2 + 10t⎤⎦| 1/2
0
⎛
⎞
= ⎝20 + 5⎠ − 0 + (30 − 15)
4
= 25.
Andrew is 25 mi from his starting point after 1 hour.
5.23 Suppose that, instead of remaining steady during the second half hour of Andrew’s outing, the wind
starts to die down according to the function v(t) = −10t + 15. In other words, the wind speed is given by
⎧20t + 5 for 0 ≤ t ≤ 1
2
v(t) = ⎨
1
⎩−10t + 15 for 2 ≤ t ≤ 1.
Under these conditions, how far from his starting point is Andrew after 1 hour?
Integrating Even and Odd Functions
We saw in Functions and Graphs that an even function is a function in which f (−x) = f (x) for all x in the
domain—that is, the graph of the curve is unchanged when x is replaced with −x. The graphs of even functions are
symmetric about the y-axis. An odd function is one in which f (−x) = − f (x) for all x in the domain, and the graph of the
function is symmetric about the origin.
Integrals of even functions, when the limits of integration are from −a to a, involve two equal areas, because they are
symmetric about the y-axis. Integrals of odd functions, when the limits of integration are similarly [−a, a], evaluate to
zero because the areas above and below the x-axis are equal.
Rule: Integrals of Even and Odd Functions
For continuous even functions such that f (−x) = f (x),
a
⌠ f (x)dx = 2∫ f (x)dx.
⌡−a
0
a
For continuous odd functions such that f (−x) = − f (x),
a
∫ −a f (x)dx = 0.
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Chapter 5 | Integration
573
Example 5.28
Integrating an Even Function
Integrate the even function
∫
2
⎛ 8
⎝3x
−2
− 2⎞⎠dx and verify that the integration formula for even functions holds.
Solution
The symmetry appears in the graphs in Figure 5.35. Graph (a) shows the region below the curve and above the
x-axis. We have to zoom in to this graph by a huge amount to see the region. Graph (b) shows the region above
the curve and below the x-axis. The signed area of this region is negative. Both views illustrate the symmetry
about the y-axis of an even function. We have
∫
2
⎛ 8
⎝3x
−2
⎛
9
|
⎞
− 2⎞⎠dx = ⎝x − 2x⎠
3
⎡ (2) 9
=⎢
⎣ 3
2
−2
⎤
⎡ (−2) 9
− 2(2)⎥ − ⎢
⎦
⎣
3
⎞ ⎛
⎞
⎛
= ⎝512 − 4⎠ − ⎝− 512 + 4⎠
3
3
⎤
− 2(−2)⎥
⎦
= 1000 .
3
To verify the integration formula for even functions, we can calculate the integral from 0 to 2 and double it, then
check to make sure we get the same answer.
∫
2
0
⎛ 8
⎝3x
⎛
9
|
⎞
− 2⎞⎠dx = ⎝x − 2x⎠
3
= 512 − 4
3
500
=
3
2
0
Since 2 · 500 = 1000 , we have verified the formula for even functions in this particular example.
3
3
Figure 5.35 Graph (a) shows the positive area between the curve and the x-axis, whereas graph (b) shows the negative area
between the curve and the x-axis. Both views show the symmetry about the y-axis.
574
Chapter 5 | Integration
Example 5.29
Integrating an Odd Function
Evaluate the definite integral of the odd function −5sin x over the interval [−π, π].
Solution
The graph is shown in Figure 5.36. We can see the symmetry about the origin by the positive area above the
x-axis over [−π, 0], and the negative area below the x-axis over [0, π]. We have
π
∫ −π−5sin xdx
= −5(−cos x)| π−π
= 5cos x| π−π
= [5cosπ] − ⎡⎣5cos(−π)⎤⎦
= −5 − (−5)
= 0.
Figure 5.36 The graph shows areas between a curve and the
x-axis for an odd function.
5.24
Integrate the function
∫
2
−2
x 4 dx.
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Chapter 5 | Integration
575
5.4 EXERCISES
Use basic integration formulas to compute the following
antiderivatives.
x − 1 ⎞⎠dx
x
220. Write an integral that quantifies the increase in the
volume of a cube when the side length doubles from s unit
to 2s units and evaluate the integral.
207.
∫ ⎛⎝
208.
⌠⎛e 2x
⌡⎝
209.
⌠dx
⌡2x
222. Write an integral that quantifies the increase in the
volume of a sphere as its radius doubles from R unit to 2R
units and evaluate the integral.
210. ⌠x −2 1 dx
223. Suppose that a particle moves along a straight line
with velocity v(t) = 4 − 2t, where 0 ≤ t ≤ 2 (in meters
⎞
− 1 e x/2⎠dx
2
⌡ x
211.
212.
∫
∫
π
0
(sin x − cos x)dx
π/2
0
(x − sin x)dx
213. Write an integral that expresses the increase in the
perimeter P(s) of a square when its side length s increases
from 2 units to 4 units and evaluate the integral.
214. Write an integral that quantifies the change in the
area A(s) = s 2 of a square when the side length doubles
from S units to 2S units and evaluate the integral.
215. A regular N-gon (an N-sided polygon with sides that
have equal length s, such as a pentagon or hexagon) has
perimeter Ns. Write an integral that expresses the increase
in perimeter of a regular N-gon when the length of each side
increases from 1 unit to 2 units and evaluate the integral.
216.
The area of a regular pentagon with side length
a > 0 is pa2 with p = 1 5 + 5 + 2 5. The Pentagon in
4
Washington, DC, has inner sides of length 360 ft and outer
sides of length 920 ft. Write an integral to express the area
of the roof of the Pentagon according to these dimensions
and evaluate this area.
217. A dodecahedron is a Platonic solid with a surface that
consists of 12 pentagons, each of equal area. By how much
does the surface area of a dodecahedron increase as the side
length of each pentagon doubles from 1 unit to 2 units?
218. An icosahedron is a Platonic solid with a surface that
consists of 20 equilateral triangles. By how much does the
surface area of an icosahedron increase as the side length of
each triangle doubles from a unit to 2a units?
219. Write an integral that quantifies the change in the
area of the surface of a cube when its side length doubles
from s unit to 2s units and evaluate the integral.
221. Write an integral that quantifies the increase in the
surface area of a sphere as its radius doubles from R unit to
2R units and evaluate the integral.
per second). Find the displacement at time t and the total
distance traveled up to t = 2.
224. Suppose that a particle moves along a straight line
with velocity defined by v(t) = t 2 − 3t − 18, where
0 ≤ t ≤ 6 (in meters per second). Find the displacement at
time t and the total distance traveled up to t = 6.
225. Suppose that a particle moves along a straight line
with velocity defined by v(t) = |2t − 6|,
where
0 ≤ t ≤ 6 (in meters per second). Find the displacement at
time t and the total distance traveled up to t = 6.
226. Suppose that a particle moves along a straight line
with acceleration defined by a(t) = t − 3,
where
0 ≤ t ≤ 6 (in meters per second). Find the velocity and
displacement at time t and the total distance traveled up to
t = 6 if v(0) = 3 and d(0) = 0.
227. A ball is thrown upward from a height of 1.5 m at
an initial speed of 40 m/sec. Acceleration resulting from
gravity is −9.8 m/sec2. Neglecting air resistance, solve for
the velocity v(t) and the height h(t) of the ball t seconds
after it is thrown and before it returns to the ground.
228. A ball is thrown upward from a height of 3 m at
an initial speed of 60 m/sec. Acceleration resulting from
gravity is −9.8 m/sec2. Neglecting air resistance, solve for
the velocity v(t) and the height h(t) of the ball t seconds
after it is thrown and before it returns to the ground.
229. The area A(t) of a circular shape is growing at a
constant rate. If the area increases from 4π units to 9π units
between times t = 2 and t = 3, find the net change in the
radius during that time.
576
Chapter 5 | Integration
230. A spherical balloon is being inflated at a constant
rate. If the volume of the balloon changes from 36π in.3 to
288π in.3 between time t = 30 and t = 60 seconds, find
the net change in the radius of the balloon during that time.
231. Water flows into a conical tank with cross-sectional
3
area πx at height x and volume πx up to height x. If
3
233. The following table lists the electrical power in
gigawatts—the rate at which energy is consumed—used in
a certain city for different hours of the day, in a typical
24-hour period, with hour 1 corresponding to midnight to 1
a.m.
Hour
Power
Hour
Power
1
28
13
48
2
25
14
49
3
24
15
49
4
23
16
50
5
24
17
50
6
27
18
50
7
29
19
46
8
32
20
43
9
34
21
42
10
39
22
40
11
42
23
37
12
46
24
34
2
water flows into the tank at a rate of 1 m3/min, find the
height of water in the tank after 5 min. Find the change in
height between 5 min and 10 min.
232. A horizontal cylindrical tank has cross-sectional area
A(x) = 4⎛⎝6x − x 2⎞⎠m 2 at height x meters above the bottom
when x ≤ 3.
a. The volume V between heights a and b is
b
∫ a A(x)dx.
Find the volume at heights between 2
m and 3 m.
b. Suppose that oil is being pumped into the tank
at a rate of 50 L/min. Using the chain rule,
dx = dx dV , at how many meters per minute is
dt dV dt
the height of oil in the tank changing, expressed in
terms of x, when the height is at x meters?
c. How long does it take to fill the tank to 3 m starting
from a fill level of 2 m?
Find the total amount of power in gigawatt-hours (gW-h)
consumed by the city in a typical 24-hour period.
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Chapter 5 | Integration
577
234. The average residential electrical power use (in
hundreds of watts) per hour is given in the following table.
Hour
Power
Hour
Power
1
8
13
12
2
3
4
5
6
7
8
9
6
14
5
15
4
16
5
17
6
18
7
19
8
20
9
21
Second
Watts
1
600
10
1200
2
500
11
1170
3
575
12
1125
4
1050
13
1100
5
925
14
1075
6
950
15
1000
7
1050
16
950
8
950
17
900
9
1100
18
780
15
17
19
18
17
16
22
16
11
10
23
13
12
11
24
11
a. Compute the average total energy used in a day in
kilowatt-hours (kWh).
b. If a ton of coal generates 1842 kWh, how long does
it take for an average residence to burn a ton of
coal?
c. Explain why the data might fit a plot of the form
⎞
Watts
14
10
⎛
Second
13
10
p(t) = 11.5 − 7.5sin⎝ πt ⎠.
12
235. The data in the following table are used to estimate
the average power output produced by Peter Sagan for each
of the last 18 sec of Stage 1 of the 2012 Tour de France.
Table 5.6 Average Power Output Source:
sportsexercisengineering.com
Estimate the net energy used in kilojoules (kJ), noting that
1W = 1 J/s, and the average power output by Sagan during
this time interval.
578
Chapter 5 | Integration
236. The data in the following table are used to estimate
the average power output produced by Peter Sagan for each
15-min interval of Stage 1 of the 2012 Tour de France.
Minutes
Watts
Minutes
Watts
15
200
165
170
237. The distribution of incomes as of 2012 in the United
States in $5000 increments is given in the following table.
The kth row denotes the percentage of households with
incomes between $5000xk and 5000xk + 4999. The row
k = 40 contains all households with income between
$200,000 and $250,000 and k = 41 accounts for all
households with income exceeding $250,000.
30
180
180
220
45
190
195
140
60
230
210
225
75
240
225
170
90
210
240
210
105
210
255
200
120
220
270
220
135
210
285
250
150
150
300
400
Table 5.7 Average Power Output Source:
sportsexercisengineering.com
Estimate the net energy used in kilojoules, noting that 1W
= 1 J/s.
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Chapter 5 | Integration
579
0
3.5
21
1.5
19
1.8
40
1
4.1
22
1.4
20
2.1
41
2
5.9
23
1.3
3
5.7
24
1.3
4
5.9
25
1.1
5
5.4
26
1.0
2.3
Table 5.8 Income
Distributions Source:
http://www.census.gov/
prod/2013pubs/p60-245.pdf
a. Estimate the percentage of U.S. households in 2012
with incomes less than $55,000.
b. What percentage of households had incomes
exceeding $85,000?
c. Plot the data and try to fit its shape to that of a
graph of the form a(x + c)e
−b(x + e)
for suitable
a, b, c.
6
5.5
27
0.75
7
5.1
28
0.8
8
4.8
29
1.0
9
4.1
30
0.6
10
4.3
31
0.6
W = ∫ F(x)dx. If Earth has mass 5.97219 × 10 24 and
radius 6371 km, compute the amount of work to elevate
a polar weather satellite of mass 1400 kg to its orbiting
altitude of 850 km above Earth.
11
3.5
32
0.5
12
3.7
33
0.5
13
3.2
34
0.4
14
3.0
35
0.3
15
2.8
36
0.3
16
2.5
37
0.3
17
2.2
38
0.2
18
2.2
39
1.8
Table 5.8 Income
Distributions Source:
http://www.census.gov/
prod/2013pubs/p60-245.pdf
238. Newton’s law of gravity states that the gravitational
force exerted by an object of mass M and one of mass
m with centers that are separated by a distance r is
F = G mM
,
r2
with
G
an
empirical
constant
G = 6.67x10 −11 m 3 /⎛⎝kg · s 2⎞⎠. The work done by a
variable force over an interval
⎡
⎣
a, b⎤⎦ is defined as
b
a
239. For a given motor vehicle, the maximum achievable
deceleration from braking is approximately 7 m/sec2 on dry
concrete. On wet asphalt, it is approximately 2.5 m/sec2.
Given that 1 mph corresponds to 0.447 m/sec, find the total
distance that a car travels in meters on dry concrete after the
brakes are applied until it comes to a complete stop if the
initial velocity is 67 mph (30 m/sec) or if the initial braking
velocity is 56 mph (25 m/sec). Find the corresponding
distances if the surface is slippery wet asphalt.
240. John is a 25-year old man who weighs 160 lb. He
burns 500 − 50t calories/hr while riding his bike for t
hours. If an oatmeal cookie has 55 cal and John eats 4t
cookies during the tth hour, how many net calories has he
lost after 3 hours riding his bike?
241. Sandra is a 25-year old woman who weighs 120
lb. She burns 300 − 50t cal/hr while walking on her
treadmill. Her caloric intake from drinking Gatorade is 100t
calories during the tth hour. What is her net decrease in
calories after walking for 3 hours?
580
Chapter 5 | Integration
242. A motor vehicle has a maximum efficiency of 33
mpg at a cruising speed of 40 mph. The efficiency drops at
a rate of 0.1 mpg/mph between 40 mph and 50 mph, and at
a rate of 0.4 mpg/mph between 50 mph and 80 mph. What
is the efficiency in miles per gallon if the car is cruising at
50 mph? What is the efficiency in miles per gallon if the car
is cruising at 80 mph? If gasoline costs $3.50/gal, what is
the cost of fuel to drive 50 mi at 40 mph, at 50 mph, and at
80 mph?
243. Although some engines are more efficient at given
a horsepower than others, on average, fuel efficiency
decreases with horsepower at a rate of 1/25 mpg/
horsepower. If a typical 50-horsepower engine has an
average fuel efficiency of 32 mpg, what is the average fuel
efficiency of an engine with the following horsepower: 150,
300, 450?
244. [T] The following table lists the 2013 schedule of
federal income tax versus taxable income.
Taxable Income
Range
The Tax Is
…
… Of the
Amount
Over
$0–$8925
10%
$0
$8925–$36,250
$892.50 +
15%
$8925
$36,250–$87,850
$4,991.25 +
25%
$36,250
$87,850–$183,250
$17,891.25
+ 28%
$87,850
$183,250–$398,350
$44,603.25
+ 33%
$183,250
$398,350–$400,000
$115,586.25
+ 35%
$398,350
> $400,000
$116,163.75
+ 39.6%
$400,000
Table 5.9 Federal Income Tax Versus Taxable
Income Source: http://www.irs.gov/pub/irs-prior/
i1040tt--2013.pdf.
Suppose that Steve just received a $10,000 raise. How
much of this raise is left after federal taxes if Steve’s salary
before receiving the raise was $40,000? If it was $90,000?
If it was $385,000?
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245. [T] The following table provides hypothetical data
regarding the level of service for a certain highway.
Highway
Speed Range
(mph)
Vehicles per
Hour per
Lane
Density
Range
(vehicles/
mi)
> 60
< 600
< 10
60–57
600–1000
10–20
57–54
1000–1500
20–30
54–46
1500–1900
30–45
46–30
1900–2100
45–70
<30
Unstable
70–200
Table 5.10
a. Plot vehicles per hour per lane on the x-axis and
highway speed on the y-axis.
b. Compute the average decrease in speed (in miles
per hour) per unit increase in congestion (vehicles
per hour per lane) as the latter increases from 600 to
1000, from 1000 to 1500, and from 1500 to 2100.
Does the decrease in miles per hour depend linearly
on the increase in vehicles per hour per lane?
c. Plot minutes per mile (60 times the reciprocal of
miles per hour) as a function of vehicles per hour
per lane. Is this function linear?
For the next two exercises use the data in the following
table, which displays bald eagle populations from 1963 to
2000 in the continental United States.
Chapter 5 | Integration
Year
Population of Breeding Pairs of
Bald Eagles
1963
487
1974
791
1981
1188
1986
1875
1992
3749
1996
5094
2000
6471
Table 5.11 Population of Breeding Bald Eagle
Pairs Source: http://www.fws.gov/Midwest/eagle/
population/chtofprs.html.
246.
[T] The graph below plots the quadratic
p(t) = 6.48t 2 − 80.3 1t + 585.69 against the data in
preceding table, normalized so that t = 0 corresponds to
1963. Estimate the average number of bald eagles per year
present for the 37 years by computing the average value of
p over [0, 37].
581
247.
[T]
3
The
graph
2
below
plots
p(t) = 0.07t + 2.42t − 25.63t + 521.23
the cubic
against the
data in the preceding table, normalized so that t = 0
corresponds to 1963. Estimate the average number of bald
eagles per year present for the 37 years by computing the
average value of p over [0, 37].
248. [T] Suppose you go on a road trip and record your
speed at every half hour, as compiled in the following
table. The best quadratic fit to the data is
q(t) = 5x 2 − 11x + 49, shown in the accompanying
graph. Integrate q to estimate the total distance driven over
the 3 hours.
Time (hr)
Speed (mph)
0 (start)
50
1
40
2
50
3
60
As a car accelerates, it does not accelerate at a constant
rate; rather, the acceleration is variable. For the following
exercises, use the following table, which contains the
acceleration measured at every second as a driver merges
onto a freeway.
582
Chapter 5 | Integration
252. [T] The number of hamburgers sold at a restaurant
throughout the day is given in the following table, with the
accompanying graph plotting the best cubic fit to the data,
b(t) = 0.12t 3 − 2.13t 3 + 12.13t + 3.91,
with t = 0
Time (sec)
Acceleration (mph/sec)
1
11.2
2
10.6
3
8.1
Hours Past Midnight
No. of Burgers Sold
4
5.4
9
3
5
0
12
28
15
20
18
30
21
45
corresponding to 9 a.m. and t = 12 corresponding to 9
p.m. Compute the average value of b(t) to estimate the
average number of hamburgers sold per hour.
249. [T] The accompanying graph plots the best quadratic
fit, a(t) = −0.70t 2 + 1.44t + 10.44, to the data from the
preceding table. Compute the average value of a(t) to
estimate the average acceleration between t = 0 and
t = 5.
250. [T] Using your acceleration equation from the
previous exercise, find the corresponding velocity
equation. Assuming the final velocity is 0 mph, find the
velocity at time t = 0.
251. [T] Using your velocity equation from the previous
exercise, find the corresponding distance equation,
assuming your initial distance is 0 mi. How far did you
travel while you accelerated your car? (Hint: You will need
to convert time units.)
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Chapter 5 | Integration
583
253. [T] An athlete runs by a motion detector, which
records her speed, as displayed in the following table. The
best linear fit to this data, ℓ(t) = −0.068t + 5.14, is
shown in the accompanying graph. Use the average value
of ℓ(t) between t = 0 and t = 40 to estimate the
runner’s average speed.
Minutes
Speed (m/sec)
0
5
10
4.8
20
3.6
30
3.0
40
2.5
584
Chapter 5 | Integration
5.5 | Substitution
Learning Objectives
5.5.1 Use substitution to evaluate indefinite integrals.
5.5.2 Use substitution to evaluate definite integrals.
The Fundamental Theorem of Calculus gave us a method to evaluate integrals without using Riemann sums. The drawback
of this method, though, is that we must be able to find an antiderivative, and this is not always easy. In this section we
examine a technique, called integration by substitution, to help us find antiderivatives. Specifically, this method helps us
find antiderivatives when the integrand is the result of a chain-rule derivative.
At first, the approach to the substitution procedure may not appear very obvious. However, it is primarily a visual task—that
is, the integrand shows you what to do; it is a matter of recognizing the form of the function. So, what are we supposed to
see? We are looking for an integrand of the form f ⎡⎣g(x)⎤⎦g′ (x)dx. For example, in the integral ⌠⎛⎝x 2 − 3⎞⎠ 2xdx, we have
⌡
3
3
2
f (x) = x , g(x) = x − 3, and g '(x) = 2x. Then,
3
f ⎡⎣g(x)⎤⎦g′ (x) = ⎛⎝x 2 − 3⎞⎠ (2x),
and we see that our integrand is in the correct form.
The method is called substitution because we substitute part of the integrand with the variable u and part of the integrand
with du. It is also referred to as change of variables because we are changing variables to obtain an expression that is easier
to work with for applying the integration rules.
Theorem 5.7: Substitution with Indefinite Integrals
Let u = g(x), , where g′ (x) is continuous over an interval, let f (x) be continuous over the corresponding range of
g, and let F(x) be an antiderivative of f (x). Then,
∫ f g(x) g′ (x)dx
⎡
⎣
⎤
⎦
= ∫ f (u)du
= F(u) + C
= F ⎛⎝g(x)⎞⎠ + C.
Proof
Let f, g, u, and F be as specified in the theorem. Then
d F(g(x)) = F′ (g(x)⎞⎠g′ (x)
dx
= f ⎡⎣g(x)⎤⎦g′ (x).
Integrating both sides with respect to x, we see that
∫ f g(x) g′ (x)dx = F g(x) + C.
⎡
⎣
⎤
⎦
⎛
⎝
⎞
⎠
If we now substitute u = g(x), and du = g '(x)dx, we get
∫ f g(x) g′ (x)dx
⎡
⎣
⎤
⎦
□
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= ∫ f (u)du
= F(u) + C
= F ⎛⎝g(x)⎞⎠ + C.
(5.19)
Chapter 5 | Integration
585
Returning to the problem we looked at originally, we let u = x 2 − 3 and then du = 2xdx. Rewrite the integral in terms of
u:
3
(2xdx) = ∫ u 3 du.
⎫
⎬
⎭
⌠⎛⎝x 2 − 3⎞⎠
⌡ u
⏟
du
Using the power rule for integrals, we have
⌠u 3 du = u 4 + C.
4
⌡
Substitute the original expression for x back into the solution:
4
⎛ 2
⎞
u 4 + C = ⎝x − 3⎠ + C.
4
4
We can generalize the procedure in the following Problem-Solving Strategy.
Problem-Solving Strategy: Integration by Substitution
1. Look carefully at the integrand and select an expression g(x) within the integrand to set equal to u. Let’s select
g(x). such that g′ (x) is also part of the integrand.
2. Substitute u = g(x) and du = g′ (x)dx. into the integral.
3. We should now be able to evaluate the integral with respect to u. If the integral can’t be evaluated we need to
go back and select a different expression to use as u.
4. Evaluate the integral in terms of u.
5. Write the result in terms of x and the expression g(x).
Example 5.30
Using Substitution to Find an Antiderivative
Use substitution to find the antiderivative of ⌠6x⎛⎝3x 2 + 4⎞⎠ dx.
4
⌡
Solution
The first step is to choose an expression for u. We choose u = 3x 2 + 4. because then du = 6xdx., and we
already have du in the integrand. Write the integral in terms of u:
⌠6x⎛3x 2 + 4⎞ dx = ∫ u 4 du.
⎝
⎠
⌡
4
Remember that du is the derivative of the expression chosen for u, regardless of what is inside the integrand. Now
we can evaluate the integral with respect to u:
586
Chapter 5 | Integration
5
= u +C
5
∫ u 4 du
=
⎛ 2
⎝3x
5
+ 4⎞⎠
+ C.
5
Analysis
We can check our answer by taking the derivative of the result of integration. We should obtain the integrand.
5
Picking a value for C of 1, we let y = 1 ⎛⎝3x 2 + 4⎞⎠ + 1. We have
5
5
y = 1 ⎛⎝3x 2 + 4⎞⎠ + 1,
5
so
4
⎛ ⎞
y′ = ⎝1 ⎠5⎛⎝3x 2 + 4⎞⎠ 6x
5
4
= 6x⎛⎝3x 2 + 4⎞⎠ .
This is exactly the expression we started with inside the integrand.
5.25
Use substitution to find the antiderivative of ⌠3x 2 ⎛⎝x 3 − 3⎞⎠ dx.
⌡
2
Sometimes we need to adjust the constants in our integral if they don’t match up exactly with the expressions we are
substituting.
Example 5.31
Using Substitution with Alteration
Use substitution to find the antiderivative of
∫z
z 2 − 5dz.
Solution
Rewrite the integral as ⌠z⎛⎝z 2 − 5⎞⎠
⌡
1/2
dz. Let u = z 2 − 5 and du = 2z dz. Now we have a problem because
du = 2z dz and the original expression has only z dz. We have to alter our expression for du or the integral in
u will be twice as large as it should be. If we multiply both sides of the du equation by 1 . we can solve this
2
problem. Thus,
u = z2 − 5
du = 2z dz
1 du = 1 (2z)dz = z dz.
2
2
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Chapter 5 | Integration
587
Write the integral in terms of u, but pull the 1 outside the integration symbol:
2
⌠z⎛z 2 − 5⎞
⎝
⎠
⌡
1/2
dz = 1 ∫ u 1/2 du.
2
Integrate the expression in u:
1 ∫ u 1/2 du = ⎛1 ⎞u 3/2 + C
⎝2 ⎠ 3
2
2
⎛ ⎞⎛ ⎞
= ⎝1 ⎠⎝2 ⎠u 3/2 + C
2 3
1
= u 3/2 + C
3
= 1 ⎛⎝z 2 − 5⎞⎠
3
5.26
3/2
+ C.
Use substitution to find the antiderivative of ⌠x 2 ⎛⎝x 3 + 5⎞⎠ dx.
9
⌡
Example 5.32
Using Substitution with Integrals of Trigonometric Functions
dt.
Use substitution to evaluate the integral ⌠ sint
3
⌡cos t
Solution
We know the derivative of cost is −sint, so we set u = cost. Then du = −sintdt. Substituting into the
integral, we have
⌠ sint dt = −⌠du .
⌡cos 3 t
⌡u 3
Evaluating the integral, we get
−⌠du3 = −∫ u −3 du
⌡u
⎛
⎞
= −⎝− 1 ⎠u −2 + C.
2
Putting the answer back in terms of t, we get
⌠ sint dt = 1 + C
⌡cos 3 t
2u 2
=
1 + C.
2cos 2 t
588
Chapter 5 | Integration
5.27
Use substitution to evaluate the integral ⌠ cost
dt.
⌡sin 2 t
Sometimes we need to manipulate an integral in ways that are more complicated than just multiplying or dividing by a
constant. We need to eliminate all the expressions within the integrand that are in terms of the original variable. When we
are done, u should be the only variable in the integrand. In some cases, this means solving for the original variable in terms
of u. This technique should become clear in the next example.
Example 5.33
Finding an Antiderivative Using u-Substitution
Use substitution to find the antiderivative of
∫
x dx.
x−1
Solution
If we let u = x − 1, then du = dx. But this does not account for the x in the numerator of the integrand. We
need to express x in terms of u. If u = x − 1, then x = u + 1. Now we can rewrite the integral in terms of u:
∫
x dx = ∫ u + 1 du
u
x−1
= ∫ u + 1 du
u
= ∫ ⎛⎝u 1/2 + u −1/2⎞⎠du.
Then we integrate in the usual way, replace u with the original expression, and factor and simplify the result.
Thus,
∫ ⎛⎝u 1/2 + u −1/2⎞⎠du
= 2 u 3/2 + 2u 1/2 + C
3
= 2 (x − 1) 3/2 + 2(x − 1) 1/2 + C
3
⎤
⎡
= (x − 1) 1/2 ⎣2 (x − 1) + 2⎦ + C
3
⎞
⎛
= (x − 1) 1/2 ⎝2 x − 2 + 6 ⎠
3 3
3
⎞
⎛
= (x − 1) 1/2 ⎝2 x + 4 ⎠
3
3
= 2 (x − 1) 1/2 (x + 2) + C.
3
5.28
Use substitution to evaluate the indefinite integral
∫ cos 3 t sint dt.
Substitution for Definite Integrals
Substitution can be used with definite integrals, too. However, using substitution to evaluate a definite integral requires a
change to the limits of integration. If we change variables in the integrand, the limits of integration change as well.
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Chapter 5 | Integration
589
Theorem 5.8: Substitution with Definite Integrals
Let u = g(x) and let g ′ be continuous over an interval ⎡⎣a, b⎤⎦, and let f be continuous over the range of u = g(x).
Then,
⌠ f g(x) g′ (x)dx = ∫
g(a)
⌡a
b
⎛
⎝
g(b)
⎞
⎠
f (u)du.
Although we will not formally prove this theorem, we justify it with some calculations here. From the substitution rule for
indefinite integrals, if F(x) is an antiderivative of f (x), we have
∫ f g(x) g′ (x)dx = F g(x) + C.
⎛
⎝
⎞
⎠
⎛
⎝
⎞
⎠
Then
b
∫ a f g(x) g′ (x)dx
⎡
⎣
⎤
⎦
(5.20)
b
= F ⎛⎝g(x)⎞⎠| xx =
=a
= F ⎛⎝g(b)⎞⎠ − F ⎛⎝g(a)⎞⎠
u = g(b)
= F(u)| u = g(a)
=∫
g(b)
g(a)
f (u)du,
and we have the desired result.
Example 5.34
Using Substitution to Evaluate a Definite Integral
1
Use substitution to evaluate ⌠ x 2 ⎛⎝1 + 2x 3⎞⎠ dx.
5
⌡0
Solution
Let u = 1 + 2x 3, so du = 6x 2 dx. Since the original function includes one factor of x2 and du = 6x 2 dx,
multiply both sides of the du equation by 1/6. Then,
du = 6x 2 dx
1 du = x 2 dx.
6
To
adjust
the
limits
of
integration,
note
that
when
x = 0, u = 1 + 2(0) = 1,
x = 1, u = 1 + 2(1) = 3. Then
1
⌠ x 2 ⎛1 + 2x 3⎞5 dx = 1 ∫ u 5 du.
⎠
⎝
6 1
⌡0
Evaluating this expression, we get
3
and
when
590
Chapter 5 | Integration
|
1 ∫ u 5 du = ⎛1 ⎞⎛u 6 ⎞
⎝6 ⎠⎝ 6 ⎠
6 1
3
⎡
6
⎣(3)
= 1
36
= 182 .
9
5.29
3
1
− (1) 6⎤⎦
0
Use substitution to evaluate the definite integral ⌠ y⎛⎝2y 2 − 3⎞⎠ dy.
5
⌡−1
Example 5.35
Using Substitution with an Exponential Function
Use substitution to evaluate
∫
1
0
xe 4x
2+3
dx.
Solution
Let u = 4x 3 + 3. Then, du = 8xdx. To adjust the limits of integration, we note that when x = 0, u = 3, and
when x = 1, u = 7. So our substitution gives
∫
1
0
xe 4x
2+3
7
dx = 1 ∫ e u du
8 3
= 1eu
8
|
7
3
3
= e −e
8
≈ 134.568.
7
5.30
Use substitution to evaluate
⌠ x 2 cos⎛π x 3⎞dx.
⎝2 ⎠
⌡0
1
Substitution may be only one of the techniques needed to evaluate a definite integral. All of the properties and rules of
integration apply independently, and trigonometric functions may need to be rewritten using a trigonometric identity before
we can apply substitution. Also, we have the option of replacing the original expression for u after we find the antiderivative,
which means that we do not have to change the limits of integration. These two approaches are shown in Example 5.36.
Example 5.36
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Chapter 5 | Integration
591
Using Substitution to Evaluate a Trigonometric Integral
Use substitution to evaluate
∫
π/2
cos 2 θ dθ.
0
Solution
Let us first use a trigonometric identity to rewrite the integral. The trig identity cos 2 θ = 1 + cos2θ allows us
2
to rewrite the integral as
∫
π/2
0
π/2
cos 2 θdθ = ⌠
⌡0
1 + cos2θ dθ.
2
Then,
⌠ ⎛1 + cos2θ ⎞dθ
⎠
⌡0 ⎝
2
π/2
⎛1
π/2
=⌠
⌡0 ⎝2
⎞
+ 1 cos2θ⎠dθ
2
π/2
π/2
= 1 ∫ dθ + ∫ cos2θdθ.
2 0
0
We can evaluate the first integral as it is, but we need to make a substitution to evaluate the second integral. Let
u = 2θ. Then, du = 2dθ, or 1 du = dθ. Also, when θ = 0, u = 0, and when θ = π/2, u = π. Expressing
2
the second integral in terms of u, we have
π/2
1 ⌠ dθ + 1 ∫ cos2θdθ = 1 ⌠ dθ + 1 ⎛1 ⎞∫ cosudu
2⌡
2⌡
2 0
2 ⎝2 ⎠ 0
0
0
π/2
π/2
=θ
2
|
θ = π/2
θ=0
π
|
u=θ
+ 1 sinu
4
u=0
⎞
⎛
= ⎝π − 0⎠ + (0 − 0) = π .
4
4
592
Chapter 5 | Integration
5.5 EXERCISES
254. Why is u-substitution referred to as change of
variable?
255.
2. If f = g ∘ h,
when reversing the chain rule,
d (g ∘ h)(x) = g′ (h(x)⎞⎠h′ (x), should you take u = g(x)
dx
or u = h(x) ?
3
267. ⌠(x − 1)⎛⎝x 2 − 2x⎞⎠ dx; u = x 2 − 2x
⌡
2
268. ⌠⎛⎝x 2 − 2x⎞⎠⎛⎝x 3 − 3x 2⎞⎠ dx; u = x 3 = 3x 2
⌡
269.
∫ cos 3 θdθ; u = sinθ
(Hint: cos 2 θ = 1 − sin 2 θ)
270.
∫ sin 3 θdθ; u = cosθ
(Hint: sin 2 θ = 1 − cos 2 θ)
In the following exercises, verify each identity using
differentiation. Then, using the indicated u-substitution,
identify f such that the integral takes the form
256.
∫x
∫ f (u)du.
x + 1dx = 2 (x + 1) 3/2 (3x − 2) + C; u = x + 1
15
257.
⌠ x 2 dx(x > 1) = 2 x − 1⎛3x 2 + 4x + 8⎞ + C; u = x − 1
⎝
⎠
15
⌡ x−1
In the following exercises, use a suitable change of
variables to determine the indefinite integral.
271.
272. ⌠t⎛⎝1 − t 2⎞⎠
⌡
258.
3/2
⌠x 4x 2 + 9dx = 1 ⎛4x 2 + 9⎞ + C; u = 4x 2 + 9
⎝
⎠
12
⌡
259.
260.
∫ x(1 − x) 99 dx
⌠ x dx = 1 4x 2 + 9 + C; u = 4x 2 + 9
4
⌡ 4x 2 + 9
x
1
⌠
dx = −
; u = 4x 2 + 9
8(4x 2 + 9)
⌡(4x 2 + 9) 2
In the following exercises, find the antiderivative using the
indicated substitution.
10
dt
273.
∫ (11x − 7) −3 dx
274.
∫ (7x − 11) 4 dx
275.
∫ cos 3 θ sinθdθ
276.
∫ sin 7 θ cosθdθ
277.
∫ cos 2 (πt)sin(πt)dt
261.
∫ (x + 1) 4 dx; u = x + 1
278.
∫ sin 2 xcos 3 xdx
262.
∫ (x − 1) 5 dx; u = x − 1
279.
∫ t sin⎛⎝t 2⎞⎠cos⎛⎝t 2⎞⎠dt
263.
∫ (2x − 3) −7 dx; u = 2x − 3
280.
∫ t 2cos 2 ⎛⎝t 3⎞⎠sin⎛⎝t 3⎞⎠dt
264.
∫ (3x − 2) −11 dx; u = 3x − 2
⌠ x dx; u = x 2 + 1
⌡ x2 + 1
281.
⌠ x2
dx
⎮⎛
2
⎞
3
⌡⎝x − 3⎠
266. ⌠ x dx; u = 1 − x 2
⌡ 1 − x2
282.
⌠ x 3 dx
⌡ 1 − x2
265.
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(Hint: sin 2 x + cos 2 x = 1)
Chapter 5 | Integration
283.
284.
285.
593
⌠ y5
dy
⎮⎛
3/2
3⎞
1
−
y
⎠
⌡⎝
π/4
297. ⌠
sinθ dθ
⌡0 cos 4 θ
∫ cosθ(1 − cosθ)
∫ ⎛⎝1 − cos 3 θ⎞⎠
10
99
In the following exercises, evaluate the indefinite integral
∫ f (x)dx
sinθdθ
Then, graph the function and the antiderivative over the
indicated interval. If possible, estimate a value of C that
would need to be added to the antiderivative to make it
cos 2 θ sinθdθ
x
286. ⌠(cosθ − 1)⎛⎝cos 2 θ − 2cosθ⎞⎠ sinθdθ
3
⌡
with constant C = 0 using u-substitution.
287. ⌠⎛⎝sin 2 θ − 2sinθ⎞⎠⎛⎝sin 3 θ − 3sin 2 θ⎞⎠ cosθdθ
⌡
3
In the following exercises, use a calculator to estimate the
area under the curve using left Riemann sums with 50
terms, then use substitution to solve for the exact answer.
equal to the definite integral F(x) = ∫ f (t)dt, with a the
a
left endpoint of the given interval.
2+x−6
298. [T]
∫ (2x + 1)e x
299. [T]
dx
∫ cos ln(2x)
x
300. [T]
⌠ 3x 2 + 2x + 1 dx over [−1, 2]
⌡ x3 + x2 + x + 4
⎞
⎠
⎛
⎝
288. [T] y = 3(1 − x) 2 over [0, 2]
dx over [−3, 2]
on [0, 2]
289. [T] y = x⎛⎝1 − x 2⎞⎠ over [−1, 2]
⎡
⎤
301. [T] ⌠ sin3x dx over ⎣− π , π ⎦
3 3
⌡cos x
290. [T] y = sin x(1 − cos x) 2 over [0, π]
302. [T]
∫ (x + 2)e −x
303. [T]
∫ 3x 2
3
291. [T] y =
⎛ 2
⎝x
x
+ 1⎞⎠
2
over [−1, 1]
In the following exercises, use a change of variables to
evaluate the definite integral.
292.
∫
1
0
x 1 − x dx
⌠ x dx
⌡0 1 + x 2
2
294.
296.
⌠ t dt
⌡0 5 + t 2
⌠ t dt
⌡0 1 + t 3
∫
π/4
0
2x 3 + 1dx over [0, 1]
304. If h(a) = h(b) in
b
∫ a g ' h(x) h(x)dx,
⎛
⎝
⎞
⎠
what can you
305. Is the substitution u = 1 − x 2 in the definite integral
2
⌠ x dx okay? If not, why not?
⌡0 1 − x 2
In the following exercises, use a change of variables to
show that each definite integral is equal to zero.
306.
∫
307.
∫
308.
∫
1
295.
dx over ⎡⎣−5, 1⎤⎦
say about the value of the integral?
2
1
293.
2 − 4x + 3
sec 2 θ tanθdθ
π
0
cos 2 (2θ)sin(2θ)dθ
π
0
1
0
t cos⎛⎝t 2⎞⎠sin⎛⎝t 2⎞⎠dt
(1 − 2t)dt
594
Chapter 5 | Integration
⌠
⎮ 1 − 2t dt
⎮ ⎛ ⎛ 1 ⎞2⎞
1+ t−
⌡0 ⎝ ⎝ 2 ⎠ ⎠
1
309.
⎛
π
⎛
310. ⌠ sin⎜⎝t −
⎝
⌡0
311.
∫
312.
∫
318. [T] The following graph is of a function of the form
f (t) = asin(nt) + bsin(mt). Estimate the coefficients a
and b, and the frequency parameters n and m. Use these
estimates to approximate
f (t)dt.
2
0
(1 − t)cos(πt)dt
3π/4
π/4
sin 2 t costdt
⎤
⎦
⎡
⎤
interval ⎣ ac , bc ⎦ for c > 0.
314. Find the area under the graph of f (t) =
⎛
⎝1
t
a
+ t 2⎞⎠
fixed, and evaluate the limit as x → ∞.
315. Find the area under the graph of g(t) =
⎛
⎝1
t
a
− t 2⎞⎠
between t = 0 and t = x, where 0 < x < 1 and a > 0
is fixed. Evaluate the limit as x → 1.
316. The area of a semicircle of radius 1 can be expressed
1
−1
319. [T] The following graph is of a function of the form
f (x) = acos(nt) + bcos(mt). Estimate the coefficients a
and b and the frequency parameters n and m. Use these
estimates to approximate
between t = 0 and t = x where a > 0 and a ≠ 1 is
∫
0
3⎞
a, b is the same as the average value of f (cx) over the
as
π
π ⎞ ⎟cos⎛t − π ⎞dt
2⎠ ⎠ ⎝ 2⎠
313. Show that the average value of f (x) over an interval
⎡
⎣
∫
1 − x 2dx. Use the substitution x = cost to
express the area of a semicircle as the integral of a
trigonometric function. You do not need to compute the
integral.
317. The area of the top half of an ellipse with a major
axis that is the x-axis from x = −1 to a and with a minor
axis that is the y-axis from y = −b to b can be written
a
2
as ⌠ b 1 − x 2 dx. Use the substitution x = acost to
a
⌡−a
express this area in terms of an integral of a trigonometric
function. You do not need to compute the integral.
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∫
π
0
f (t)dt.
Chapter 5 | Integration
595
5.6 | Integrals Involving Exponential and Logarithmic
Functions
Learning Objectives
5.6.1 Integrate functions involving exponential functions.
5.6.2 Integrate functions involving logarithmic functions.
Exponential and logarithmic functions are used to model population growth, cell growth, and financial growth, as well as
depreciation, radioactive decay, and resource consumption, to name only a few applications. In this section, we explore
integration involving exponential and logarithmic functions.
Integrals of Exponential Functions
The exponential function is perhaps the most efficient function in terms of the operations of calculus. The exponential
function, y = e x, is its own derivative and its own integral.
Rule: Integrals of Exponential Functions
Exponential functions can be integrated using the following formulas.
∫ e x dx
= ex + C
∫ a x dx
=
(5.21)
ax + C
lna
Example 5.37
Finding an Antiderivative of an Exponential Function
Find the antiderivative of the exponential function e−x.
Solution
Use substitution, setting u = −x, and then du = −1dx. Multiply the du equation by −1, so you now have
−du = dx. Then,
∫ e −xdx
= −∫ e u du
= −e u + C
= −e −x + C.
5.31
3
Find the antiderivative of the function using substitution: x 2 e −2x .
A common mistake when dealing with exponential expressions is treating the exponent on e the same way we treat
exponents in polynomial expressions. We cannot use the power rule for the exponent on e. This can be especially confusing
when we have both exponentials and polynomials in the same expression, as in the previous checkpoint. In these cases, we
should always double-check to make sure we’re using the right rules for the functions we’re integrating.
596
Chapter 5 | Integration
Example 5.38
Square Root of an Exponential Function
Find the antiderivative of the exponential function e x 1 + e x.
Solution
First rewrite the problem using a rational exponent:
∫ ex
1 + e xdx = ∫ e x (1 + e x) 1/2 dx.
Using substitution, choose u = 1 + e x.u = 1 + e x. Then, du = e x dx. We have (Figure 5.37)
∫ e x (1 + e x) 1/2 dx = ∫ u 1/2 du.
Then
⌠u 1/2 du = u 3/2 + C = 2 u 3/2 + C = 2 (1 + e x) 3/2 + C.
3/2
3
3
⌡
Figure 5.37 The graph shows an exponential function times
the square root of an exponential function.
5.32
Find the antiderivative of e x (3e x − 2) 2.
Example 5.39
Using Substitution with an Exponential Function
Use substitution to evaluate the indefinite integral
∫ 3x 2 e 2x
3
dx.
Solution
Here we choose to let u equal the expression in the exponent on e. Let u = 2x 3 and du = 6x 2 dx.. Again, du
is off by a constant multiplier; the original function contains a factor of 3x2, not 6x2. Multiply both sides of the
equation by 1 so that the integrand in u equals the integrand in x. Thus,
2
∫ 3x 2 e 2x
3
dx = 1 ∫ e u du.
2
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Chapter 5 | Integration
597
Integrate the expression in u and then substitute the original expression in x back into the u integral:
1 ∫ e udu = 1 e u + C = 1 e 2x 3 + C.
2
2
2
5.33
Evaluate the indefinite integral
∫ 2x 3 e x
4
dx.
As mentioned at the beginning of this section, exponential functions are used in many real-life applications. The number e is
often associated with compounded or accelerating growth, as we have seen in earlier sections about the derivative. Although
the derivative represents a rate of change or a growth rate, the integral represents the total change or the total growth. Let’s
look at an example in which integration of an exponential function solves a common business application.
A price–demand function tells us the relationship between the quantity of a product demanded and the price of the product.
In general, price decreases as quantity demanded increases. The marginal price–demand function is the derivative of the
price–demand function and it tells us how fast the price changes at a given level of production. These functions are used in
business to determine the price–elasticity of demand, and to help companies determine whether changing production levels
would be profitable.
Example 5.40
Finding a Price–Demand Equation
Find the price–demand equation for a particular brand of toothpaste at a supermarket chain when the demand is
50 tubes per week at $2.35 per tube, given that the marginal price—demand function, p′ (x), for x number of
tubes per week, is given as
p '(x) = −0.015e −0.01x.
If the supermarket chain sells 100 tubes per week, what price should it set?
Solution
To find the price–demand equation, integrate the marginal price–demand function. First find the antiderivative,
then look at the particulars. Thus,
p(x) = ∫ −0.015e −0.01x dx
= −0.015∫ e −0.01xdx.
Using substitution, let u = −0.01x and du = −0.01dx. Then, divide both sides of the du equation by −0.01.
This gives
−0.015 ∫ e u du = 1.5∫ e u du
−0.01
= 1.5e u + C
= 1.5e −0.01x + C.
The next step is to solve for C. We know that when the price is $2.35 per tube, the demand is 50 tubes per week.
This means
598
Chapter 5 | Integration
p(50) = 1.5e
−0.01(50)
+C
= 2.35.
Now, just solve for C:
C = 2.35 − 1.5e −0.5
= 2.35 − 0.91
= 1.44.
Thus,
p(x) = 1.5e −0.01x + 1.44.
If the supermarket sells 100 tubes of toothpaste per week, the price would be
p(100) = 1.5e
−0.01(100)
+ 1.44 = 1.5e −1 + 1.44 ≈ 1.99.
The supermarket should charge $1.99 per tube if it is selling 100 tubes per week.
Example 5.41
Evaluating a Definite Integral Involving an Exponential Function
Evaluate the definite integral
∫
2
1
e 1 − x dx.
Solution
Again, substitution is the method to use. Let
∫ e 1 − x dx = −∫ e u du.
u = 1 − x,
so
du = −1dx or −du = dx. Then
Next, change the limits of integration. Using the equation u = 1 − x, we have
u = 1 − (1) = 0
u = 1 − (2) = −1.
The integral then becomes
∫
2
1
e 1 − x dx = −∫
=∫
=
0
−1
0
e u du
e u du
−1
u0
e −1
0 ⎛ −1⎞
|
= e − ⎝e
= −e
See Figure 5.38.
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−1
⎠
+ 1.
Chapter 5 | Integration
599
Figure 5.38 The indicated area can be calculated by
evaluating a definite integral using substitution.
5.34
Evaluate
∫
2
0
e 2x dx.
Example 5.42
Growth of Bacteria in a Culture
Suppose the rate of growth of bacteria in a Petri dish is given by q(t) = 3 t, where t is given in hours and q(t)
is given in thousands of bacteria per hour. If a culture starts with 10,000 bacteria, find a function Q(t) that gives
the number of bacteria in the Petri dish at any time t. How many bacteria are in the dish after 2 hours?
Solution
We have
t
Q(t) = ⌠3 t dt = 3 + C.
ln3
⌡
Then, at t = 0 we have Q(0) = 10 = 1 + C, so C ≈ 9.090 and we get
ln3
t
Q(t) = 3 + 9.090.
ln3
At time t = 2, we have
2
Q(2) = 3 + 9.090
ln3
= 17.282.
After 2 hours, there are 17,282 bacteria in the dish.
5.35
From Example 5.42, suppose the bacteria grow at a rate of q(t) = 2 t. Assume the culture still starts
with 10,000 bacteria. Find Q(t). How many bacteria are in the dish after 3 hours?
600
Chapter 5 | Integration
Example 5.43
Fruit Fly Population Growth
Suppose a population of fruit flies increases at a rate of g(t) = 2e 0.02t, in flies per day. If the initial population
of fruit flies is 100 flies, how many flies are in the population after 10 days?
Solution
Let G(t) represent the number of flies in the population at time t. Applying the net change theorem, we have
G(10) = G(0) + ∫
10
0
2e 0.02t dt
|
⎤
⎡
= 100 + ⎣ 2 e 0.02t⎦ 0
0.02
= 100 +
|
10
10
⎡
0.02t⎤
⎣100e
⎦0
= 100 + 100e 0.2 − 100
≈ 122.
There are 122 flies in the population after 10 days.
5.36
Suppose the rate of growth of the fly population is given by g(t) = e 0.01t, and the initial fly population
is 100 flies. How many flies are in the population after 15 days?
Example 5.44
Evaluating a Definite Integral Using Substitution
2
1/x
Evaluate the definite integral using substitution: ⌠ e 2 dx.
⌡1 x
Solution
This problem requires some rewriting to simplify applying the properties. First, rewrite the exponent on e as a
power of x, then bring the x2 in the denominator up to the numerator using a negative exponent. We have
2
⌠ e 1/x dx = ∫ e x −1 x −2 dx.
⌡1 x 2
1
2
Let u = x −1, the exponent on e. Then
du = −x −2 dx
−du = x −2 dx.
Bringing the negative sign outside the integral sign, the problem now reads
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Chapter 5 | Integration
601
−∫ e u du.
Next, change the limits of integration:
u = (1) −1 = 1
u = (2) −1 = 1 .
2
Notice that now the limits begin with the larger number, meaning we must multiply by −1 and interchange the
limits. Thus,
−∫
1/2
1
e u du = ∫
1
e u du
1/2
= eu |1
1/2
1/2
=e−e
= e − e.
5.37
2
−2
Evaluate the definite integral using substitution: ⌠ 13 e 4x dx.
⌡1 x
Integrals Involving Logarithmic Functions
Integrating functions of the form f (x) = x −1 result in the absolute value of the natural log function, as shown in the
following rule. Integral formulas for other logarithmic functions, such as f (x) = ln x and f (x) = log a x, are also included
in the rule.
Rule: Integration Formulas Involving Logarithmic Functions
The following formulas can be used to evaluate integrals involving logarithmic functions.
∫ x −1 dx
∫ ln x dx
∫ log ax dx
= ln|x| + C
= x ln x − x + C = x(ln x − 1) + C
=
x (ln x − 1) + C
lna
Example 5.45
Finding an Antiderivative Involving ln x
Find the antiderivative of the function
3 .
x − 10
(5.22)
602
Chapter 5 | Integration
Solution
First factor the 3 outside the integral symbol. Then use the u−1 rule. Thus,
⌠ 3 dx
⌡x − 10
= 3⌠ 1 dx
⌡x − 10
= 3∫ du
u
= 3ln|u| + C
= 3ln|x − 10| + C, x ≠ 10.
See Figure 5.39.
Figure 5.39 The domain of this function is x ≠ 10.
5.38
Find the antiderivative of
1 .
x+2
Example 5.46
Finding an Antiderivative of a Rational Function
3
Find the antiderivative of 2x4 + 3x2 .
x + 3x
Solution
This can be rewritten as
⌠⎛2x 3 + 3x⎞⎛x 4 + 3x 2⎞
⎠
⎠⎝
⎝
⌡
−1
dx. Use substitution. Let
du = 4x 3 + 6x. Alter du by factoring out the 2. Thus,
du =
⎛ 3
⎝4x
+ 6x⎞⎠dx
⎛ 3
⎝2x
+ 3x⎞⎠dx.
= 2⎛⎝2x 3 + 3x⎞⎠dx
1 du =
2
Rewrite the integrand in u:
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u = x 4 + 3x 2,
then
Chapter 5 | Integration
603
⌠⎛2x 3 + 3x⎞⎛x 4 + 3x 2⎞
⎠
⎠⎝
⎝
⌡
−1
dx = 1 ∫ u −1du.
2
Then we have
1 ∫ u −1 du = 1 ln|u| + C
2
2
|
|
= 1 ln x 4 + 3x 2 + C.
2
Example 5.47
Finding an Antiderivative of a Logarithmic Function
Find the antiderivative of the log function log 2 x.
Solution
Follow the format in the formula listed in the rule on integration formulas involving logarithmic functions. Based
on this format, we have
x (ln x − 1) + C.
∫ log 2 xdx = ln2
5.39
Find the antiderivative of log 3 x.
Example 5.48 is a definite integral of a trigonometric function. With trigonometric functions, we often have to apply a
trigonometric property or an identity before we can move forward. Finding the right form of the integrand is usually the key
to a smooth integration.
Example 5.48
Evaluating a Definite Integral
π/2
Find the definite integral of ⌠
⌡0
sin x dx.
1 + cos x
Solution
We need substitution to evaluate this problem. Let u = 1 + cos x, , so du = −sin x dx. Rewrite the integral in
terms of u, changing the limits of integration as well. Thus,
u = 1 + cos(0) = 2
⎛ ⎞
u = 1 + cos⎝π ⎠ = 1.
2
604
Chapter 5 | Integration
Then
π/2
sin x
⌠
⌡0 1 + cos x
1
= −∫ u −1 du
2
2
= ∫ u −1 du
1
= ln|u|| 21
= [ln2 − ln1]
= ln2.
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Chapter 5 | Integration
605
5.6 EXERCISES
In the following exercises, compute each indefinite
integral.
337.
∫ x 2e −x
3
dx
320.
∫ e 2xdx
338.
∫ e sin x cos xdx
321.
∫ e −3xdx
339.
∫ e tan xsec 2 xdx
322.
∫ 2 xdx
340.
∫ e ln x dx
x
323.
∫ 3 −xdx
341. ⌠e
324.
⌠ 1 dx
⌡2x
In the following exercises, verify by differentiation that
325.
∫ 2x dx
⌡ 1−t
then
use
appropriate
changes of variables to compute the integral.
⌡x
∫
dt
∫ ln x dx = x(ln x − 1) + C,
326. ⌠ 12 dx
327.
ln(1 − t)
1 dx
x
In the following exercises, find each indefinite integral by
using appropriate substitutions.
328.
∫ lnxx dx
329.
⌠ dx
⌡x(ln x) 2
(Hint: ⌠ln xdx = 1 ∫ xln⎛⎝x 2⎞⎠dx)
⌡
2
342.
∫ ln xdx
343.
∫ x 2ln 2 x dx
344. ⌠ln2x dx (Hint: Set u = 1x .)
⌡x
345.
∫ lnxx dx
(Hint: Set u = x.)
346. Write an integral to express the area under the graph
of y = 1t from t = 1 to ex and evaluate the integral.
330. ⌠ dx (x > 1)
347. Write an integral to express the area under the graph
of y = e t between t = 0 and t = ln x, and evaluate the
dx
331. ⌠
⌡x ln x ln(ln x)
integral.
⌡x ln x
332.
∫ tanθ dθ
333.
− x sin x dx
∫ cos xx cos
x
ln(sin x)
334. ⌠
dx
⌡ tan x
335.
∫ ln(cos x)tan xdx
336.
∫ xe −x
2
dx
In the following exercises, use appropriate substitutions
to express the trigonometric integrals in terms of
compositions with logarithms.
348.
∫ tan(2x)dx
sin(3x) − cos(3x)
349. ⌠
dx
⌡sin(3x) + cos(3x)
350.
⌠x sin⎝x ⎠dx
⎛ 2⎞
⌡ cos⎝x ⎠
351.
∫ x csc⎛⎝x 2⎞⎠dx
⎛ 2⎞
606
Chapter 5 | Integration
352.
∫ ln(cos x)tan x dx
368. [T] y = ln(x) over [1, 2]
353.
∫ ln(csc x)cot xdx
369. [T] y = 2 x + 1
over [0, 1]
x + 2x + 6
354.
⌠e − e dx
⌡e x + e −x
370. [T] y = 2 x over [−1, 0]
−x
x
In the following exercises, evaluate the definite integral.
2
355. ⌠ 1 + 2x 2+ x 3 dx
⌡ 3x + 3x + x
2
1
371. [T] y = −2 −x over [0, 1]
In the following exercises, f (x) ≥ 0 for a ≤ x ≤ b. Find
the area under the graph of f (x) between the given values
a and b by integrating.
356.
∫
tan x dx
372.
f (x) =
log 10 (x)
; a = 10, b = 100
x
373.
f (x) =
357.
π/3
⌠ sin x − cos x dx
⌡0 sin x + cos x
log 2 (x)
x ; a = 32, b = 64
374.
f (x) = 2 −x; a = 1, b = 2
375.
f (x) = 2 −x; a = 3, b = 4
376.
Find the area under the graph of the function
358.
359.
∫
∫
π/4
0
π/2
csc xdx
π/6
π/3
π/4
In the following exercises, integrate using the indicated
substitution.
360.
∫ x −x100 dx; u = x − 100
y−1
361. ⌠
dy; u = y + 1
⌡y + 1
2
362. ⌠ 1 − x 3 dx; u = 3x − x 3
⌡3x − x
363. ⌠sin x + cos x dx; u = sin x − cos x
⌡sin x − cos x
364.
∫ e 2x
365.
⌠ln(x) 1 − (ln x) 2 dx; u = ln x
x
⌡
1 − e 2xdx; u = e 2x
In the following exercises, does the right-endpoint
approximation overestimate or underestimate the exact
area? Calculate the right endpoint estimate R50 and solve
for the exact area.
366. [T] y = e x over [0, 1]
367. [T] y = e
−x
2
f (x) = xe −x between x = 0 and x = 5.
cot xdx
over [0, 1]
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2
377. Compute the integral of f (x) = xe −x and find the
smallest value of N such that the area under the graph
f (x) = xe −x
2
between x = N and x = N + 10 is, at
most, 0.01.
378. Find the limit, as N tends to infinity, of the area under
2
the graph of f (x) = xe −x between x = 0 and x = 5.
379. Show that
b
∫ a dtt = ∫
1/a
1/b
dt when 0 < a ≤ b.
t
380. Suppose that f (x) > 0 for all x and that f and g are
differentiable. Use the identity f g = e
g ln f
and the chain
g
rule to find the derivative of f .
381. Use the previous exercise to find the antiderivative of
h(x) = x x (1 + ln x) and evaluate
∫
3
2
x x (1 + ln x)dx.
382. Show that if c > 0, then the integral of 1/x from
ac to bc (0 < a < b) is the same as the integral of 1/x
from a to b.
The following exercises are intended to derive the
fundamental properties of the natural log starting from the
Chapter 5 | Integration
607
x
definition ln(x) = ∫ dt
t , using properties of the definite
1
integral and making no further assumptions.
x
383. Use the identity ln(x) = ∫ dt
t to derive the identity
ln⎛⎝1x ⎞⎠ =
1
−ln x.
384. Use a change of variable in the integral
∫
show that ln xy = ln x + ln y for x, y > 0.
xy
1
1 dt to
t
x
385. Use the identity ln x = ∫ dt
x to show that ln(x)
1
is an increasing function of x on [0, ∞),
and use the
previous exercises to show that the range of ln(x) is
(−∞, ∞). Without any further assumptions, conclude that
ln(x) has an inverse function defined on (−∞, ∞).
386. Pretend, for the moment, that we do not know that
e x is the inverse function of ln(x), but keep in mind
that ln(x) has an inverse function defined on (−∞, ∞).
Call it E. Use the identity ln xy = ln x + ln y to deduce that
E(a + b) = E(a)E(b) for any real numbers a, b.
387. Pretend, for the moment, that we do not know that
e x is the inverse function of ln x, but keep in mind that
ln x has an inverse function defined on (−∞, ∞). Call it
E. Show that E '(t) = E(t).
x
388. The sine integral, defined as S(x) = ∫ sint
t dt is
0
an important quantity in engineering. Although it does not
have a simple closed formula, it is possible to estimate
its behavior for large x. Show that for
k ≥ 1, |S(2πk) − S⎛⎝2π(k + 1)⎞⎠| ≤
1
.
k(2k + 1)π
(Hint: sin(t + π) = −sint)
389. [T] The normal distribution in probability is given
by p(x) =
1 e −(x − µ)
σ 2π
2 /2σ 2
, where σ is the standard
deviation and μ is the average. The standard normal
distribution in probability, p s,
corresponds to
µ = 0 and σ = 1. Compute the left endpoint estimates
1
R 10 and R 100 of ⌠
⌡−1
1 e −x 2/2 dx.
2π
390.
[T]
Compute
5
the
right
1 e −(x − 1)
R 50 and R 100 of ⌠
⌡−3 2 2π
endpoint
2 /8
.
estimates
608
Chapter 5 | Integration
5.7 | Integrals Resulting in Inverse Trigonometric
Functions
Learning Objectives
5.7.1 Integrate functions resulting in inverse trigonometric functions
In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions
before. Recall from Functions and Graphs that trigonometric functions are not one-to-one unless the domains are
restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions
into account. Also in Derivatives, we developed formulas for derivatives of inverse trigonometric functions. The formulas
developed there give rise directly to integration formulas involving inverse trigonometric functions.
Integrals that Result in Inverse Sine Functions
Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to
evaluate the integrals. We prove the formula for the inverse sine integral.
Rule: Integration Formulas Resulting in Inverse Trigonometric Functions
The following integration formulas yield inverse trigonometric functions:
1.
⌠ du = sin −1 u + C
a
⌡ a2 − u2
(5.23)
⌠ du = 1 tan −1 u + C
a
⌡a 2 + u 2 a
(5.24)
⌠ du
= 1a sec −1 ua + C
2
2
⌡u u − a
(5.25)
2.
3.
Proof
Let y = sin −1 ax . Then asin y = x. Now let’s use implicit differentiation. We obtain
d ⎛⎝asin y⎞⎠ = d (x)
dx
dx
dy
acos y
= 1
dx
dy
1 .
= acos
y
dx
For
− π ≤ y ≤ π , cos y ≥ 0.
2
2
Thus,
applying
the
Pythagorean
cos y = 1 = sin 2 y. This gives
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identity
sin 2 y + cos 2 y = 1,
we
have
Chapter 5 | Integration
609
1
acos y =
1
a 1 − sin 2 y
1
a − a 2 sin 2 y
1
=
.
2
a − x2
=
2
Then for −a ≤ x ≤ a, we have
⌠ 1 du = sin −1⎛⎝u ⎞⎠ + C.
a
⌡ a2 − u2
□
Example 5.49
Evaluating a Definite Integral Using Inverse Trigonometric Functions
1
dx .
Evaluate the definite integral ⌠
⌡0 1 − x 2
Solution
We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse
trigonometric functions, and then evaluate the definite integral. We have
1
1
⌠ dx
= sin −1 x
0
⌡0 1 − x 2
|
= sin −1 1 − sin −1 0
= π −0
2
= π.
2
5.40
Find the antiderivative of ⌠
dx
.
⌡ 1 − 16x 2
Example 5.50
Finding an Antiderivative Involving an Inverse Trigonometric Function
Evaluate the integral
Solution
⌠ dx .
⌡ 4 − 9x 2
610
Chapter 5 | Integration
Substitute u = 3x. Then du = 3dx and we have
⌠ dx = 1 ⌠ du .
⌡ 4 − 9x 2 3 ⌡ 4 − u 2
Applying the formula with a = 2, we obtain
⌠ dx
= 1 ⌠ du
3⌡ 4 − u2
⌡ 4 − 9x 2
⎛ ⎞
= 1 sin −1 ⎝u ⎠ + C
3
2
−1 ⎛3x ⎞
1
= sin ⎝ ⎠ + C.
3
2
5.41
Find the indefinite integral using an inverse trigonometric function and substitution for ⌠ dx .
⌡ 9 − x2
Example 5.51
Evaluating a Definite Integral
3/2
Evaluate the definite integral ⌠
⌡0
du .
1 − u2
Solution
The format of the problem matches the inverse sine formula. Thus,
3/2
⌠
⌡0
3/2
du
= sin −1 u
2
0
1−u
⎡
⎛ ⎞⎤
= ⎣sin −1 ⎝ 3 ⎠⎦ − ⎡⎣sin −1 (0)⎤⎦
2
π
= .
3
|
Integrals Resulting in Other Inverse Trigonometric Functions
There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration
formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use.
The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the
integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close
this section, we examine one more formula: the integral resulting in the inverse tangent function.
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Chapter 5 | Integration
611
Example 5.52
Finding an Antiderivative Involving the Inverse Tangent Function
Find an antiderivative of ⌠
1 dx.
⌡1 + 4x 2
Solution
Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse
trigonometric functions, the integrand looks similar to the formula for tan −1 u + C. So we use substitution,
letting u = 2x, then du = 2dx and 1/2du = dx. Then, we have
1 ⌠ 1 du = 1 tan −1 u + C = 1 tan −1 (2x) + C.
2 ⌡1 + u 2
2
2
5.42
Use substitution to find the antiderivative of ⌠ dx 2 .
⌡25 + 4x
Example 5.53
Applying the Integration Formulas
Find the antiderivative of ⌠
1 dx.
⌡9 + x 2
Solution
Apply the formula with a = 3. Then,
⌠ dx = 1 tan −1 ⎛ x ⎞ + C.
⎝3 ⎠
⌡9 + x 2 3
5.43
Find the antiderivative of ⌠ dx 2 .
⌡16 + x
Example 5.54
Evaluating a Definite Integral
Evaluate the definite integral ⌠
⌡
3
dx .
1
+
x2
3/3
612
Chapter 5 | Integration
Solution
Use the formula for the inverse tangent. We have
⌠
⌡
5.44
3
dx
= tan −1 x
2
1
+
x
3/3
|
3
3/3
⎡
⎛ ⎞⎤
= ⎡⎣tan −1 ⎛⎝ 3⎞⎠⎤⎦ − ⎣tan −1 ⎝ 3 ⎠⎦
3
π
= .
6
2
Evaluate the definite integral ⌠ dx 2 .
⌡0 4 + x
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Chapter 5 | Integration
613
5.7 EXERCISES
In the following exercises, evaluate each integral in terms
of an inverse trigonometric function.
3/2
391. ⌠
⌡0
dx
1 − x2
1/2
392.
dx
⌠
⌡−1/2 1 − x 2
1
393.
dx
⌠
⌡ 3 1 + x2
394. ⌠
3
dx
⌡1/ 3 1 + x 2
2
395.
dx
⌠
⌡1 |x| x 2 − 1
2/ 3
396.
dx
⌠
⌡1 |x| x 2 − 1
In the following exercises, find each indefinite integral,
using appropriate substitutions.
397.
⌠ dx
⌡ 9 − x2
398.
⌠ dx
⌡ 1 − 16x 2
399. ⌠ dx 2
⌡9 + x
dx
400. ⌠
⌡25 + 16x 2
401.
402.
⌠ dx
⌡|x| x 2 − 9
dx
⌠
⌡|x| 4x 2 − 16
403.
Explain
the
relationship
−cos −1 t + C = ⌠ dt = sin −1 t + C. Is it true, in
⌡ 1 − t2
general, that cos −1 t = −sin −1 t ?
404.
Explain
the
relationship
sec −1 t + C = ⌠ dt
= −csc −1 t + C. Is it true, in
⌡|t| t 2 − 1
general, that sec −1 t = −csc −1 t ?
405. Explain what is wrong with the following integral:
2
⌠ dt .
⌡1 1 − t 2
406. Explain what is wrong with the following integral:
1
dt .
⌠
⌡−1 |t| t 2 − 1
In the following exercises, solve for the antiderivative
of f with C = 0,
∫f
then use a calculator to graph f and
the antiderivative over the given interval ⎡⎣a, b⎤⎦. Identify a
value of C such that adding C to the antiderivative recovers
x
the definite integral F(x) = ∫ f (t)dt.
a
407. [T]
⌠ 1 dx over [−3, 3]
⌡ 9 − x2
408. [T] ⌠
9 dx over ⎡⎣−6, 6⎤⎦
⌡9 + x 2
409. [T] ⌠ cos x2 dx over ⎡⎣−6, 6⎤⎦
⌡4 + sin x
410. [T] ⌠
e x dx over ⎡⎣−6, 6⎤⎦
⌡1 + e 2x
In the following exercises, compute the antiderivative using
appropriate substitutions.
411.
⌠sin −1 tdt
⌡ 1 − t2
412.
dt
⌠
−1
⌡sin t 1 − t 2
614
Chapter 5 | Integration
tan
413. ⌠
−1
(2t)
⌡ 1 + 4t 2
dt
414.
⌠ttan ⎝t
4
⌡ 1+t
415.
⌠sec ⎝2 ⎠ dt
⌡|t| t 2 − 4
416.
⌠tsec ⎝t ⎠dt
⌡t 2 t 4 − 1
−1 ⎛ 2⎞
⎠
dt
426.
⌠ dt
⎛
2 ⎞
⌡t⎝1 + ln t⎠
427.
⌠cos (2t) dt
⌡ 1 − 4t 2
428.
t
−1 ⎛ t⎞
⌠e cos ⎝e ⎠dt
⌡ 1 − e 2t
−1 ⎛ t ⎞
−1 ⎛ 2⎞
In the following exercises, compute each definite integral.
∫f
429.
⌠
⌡0
430.
⌠ tan⎝cos t⎠dt
⌡1/4 1 − t 2
431.
⌠
⌡0
sin⎛⎝tan −1 t⎞⎠
432.
⌠
⌡0
cos⎛⎝tan −1 t⎞⎠
433.
For A > 0,
1/2
In the following exercises, use a calculator to graph the
antiderivative
with C = 0 over the given interval
⎡
⎣
a, b⎤⎦. Approximate a value of C, if possible, such that
adding C to the antiderivative gives the same value as the
x
a
⌠ 1 dx over 2, 6
⌡x x 2 − 4
⎡
⎣
1/2
⎤
⎦
418. [T] ⌠
1
dx over ⎡⎣0, 6⎤⎦
⌡(2x + 2) x
(sin x + x cos x)
419. [T] ⌠
dx over ⎡⎣−6, 6⎤⎦
2
2
⌡ 1 + x sin x
420. [T]
⌠ 2e −2x dx over [0, 2]
⌡ 1 − e −4x
421. [T] ⌠
1
over [0, 2]
⌡x + xln 2 x
422. [T]
⌠ sin −1 x over [−1, 1]
⌡ 1 − x2
In the following exercises, compute each integral using
appropriate substitutions.
423.
⌠ e x dt
⌡ 1 − e 2t
424. ⌠
e t dt
⌡1 + e 2t
425.
⌠ dt
⌡t 1 − ln 2 t
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tan⎛⎝sin −1 t⎞⎠
1 − t2
1/2
definite integral F(x) = ∫ f (t)dt.
417. [T]
−1
⎛
−1 ⎞
1 + t2
1/2
dt
dt
1 + t2
dt
A
compute I(A) = ⌠
dt
and
⌡−A 1 + t 2
1
evaluate a lim
I(A), the area under the graph of
→∞
1 + t2
on [−∞, ∞].
B
434. For 1 < B < ∞, compute I(B) = ⌠
evaluate
lim I(B),
B→∞
dt
and
⌡1 t t − 1
2
the area under the graph of
1
over [1, ∞).
t t2 − 1
435. Use the substitution u = 2 cot x and the identity
1 + cot 2 x = csc 2 x to evaluate ⌠ dx 2 . (Hint:
⌡1 + cos x
Multiply the top and bottom of the integrand by csc 2 x.)
Chapter 5 | Integration
615
436. [T] Approximate the points at which the graphs of
f (x) = 2x 2 − 1 and g(x) = ⎛⎝1 + 4x 2⎞⎠
−3/2
intersect, and
approximate the area between their graphs accurate to three
decimal places.
437. 47. [T] Approximate the points at which the graphs
of f (x) = x 2 − 1 and f (x) = x 2 − 1 intersect, and
approximate the area between their graphs accurate to three
decimal places.
438.
x
⌠
⌡0
Use
the
following
graph
1 − t dt = 1 x 1 − x 2 + 1 sin −1 x.
2
2
2
to
prove
that
616
Chapter 5 | Integration
CHAPTER 5 REVIEW
KEY TERMS
average value of a function (or fave) the average value of a function on an interval can be found by calculating the
definite integral of the function and dividing that value by the length of the interval
change of variables the substitution of a variable, such as u, for an expression in the integrand
definite integral a primary operation of calculus; the area between the curve and the x-axis over a given interval is a
definite integral
fundamental theorem of calculus the theorem, central to the entire development of calculus, that establishes the
relationship between differentiation and integration
fundamental theorem of calculus, part 1 uses a definite integral to define an antiderivative of a function
fundamental theorem of calculus, part 2 (also, evaluation theorem) we can evaluate a definite integral by
evaluating the antiderivative of the integrand at the endpoints of the interval and subtracting
integrable function a function is integrable if the limit defining the integral exists; in other words, if the limit of the
Riemann sums as n goes to infinity exists
integrand the function to the right of the integration symbol; the integrand includes the function being integrated
integration by substitution a technique for integration that allows integration of functions that are the result of a
chain-rule derivative
left-endpoint approximation an approximation of the area under a curve computed by using the left endpoint of each
subinterval to calculate the height of the vertical sides of each rectangle
limits of integration these values appear near the top and bottom of the integral sign and define the interval over which
the function should be integrated
lower sum a sum obtained by using the minimum value of f (x) on each subinterval
mean value theorem for integrals guarantees that a point c exists such that f (c) is equal to the average value of the
function
net change theorem if we know the rate of change of a quantity, the net change theorem says the future quantity is
equal to the initial quantity plus the integral of the rate of change of the quantity
net signed area the area between a function and the x-axis such that the area below the x-axis is subtracted from the area
above the x-axis; the result is the same as the definite integral of the function
partition a set of points that divides an interval into subintervals
regular partition a partition in which the subintervals all have the same width
riemann sum
an estimate of the area under the curve of the form A ≈
n
∑
i=1
f (x*i )Δx
right-endpoint approximation the right-endpoint approximation is an approximation of the area of the rectangles
under a curve using the right endpoint of each subinterval to construct the vertical sides of each rectangle
sigma notation (also, summation notation) the Greek letter sigma (Σ) indicates addition of the values; the values of the
index above and below the sigma indicate where to begin the summation and where to end it
total area total area between a function and the x-axis is calculated by adding the area above the x-axis and the area
below the x-axis; the result is the same as the definite integral of the absolute value of the function
upper sum a sum obtained by using the maximum value of f (x) on each subinterval
variable of integration indicates which variable you are integrating with respect to; if it is x, then the function in the
integrand is followed by dx
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Chapter 5 | Integration
617
KEY EQUATIONS
• Properties of Sigma Notation
n
∑ c = nc
i=1
n
n
∑ ca i = c ∑ a i
i=1
n
i=1
∑
⎛
⎝
∑
⎛
⎝
i=1
n
i=1
n
n
n
i=1
n
i=1
n
∑ ai + ∑ bi
a i + b i⎞⎠ =
∑ ai − ∑ bi
a i − b i⎞⎠ =
i=1
m
i=1
n
∑ ai = ∑ ai + ∑
i=1
i=m+1
i=1
ai
• Sums and Powers of Integers
n
∑ i = 1 + 2 + ⋯ + n = n(n 2+ 1)
i=1
n
+ 1)
∑ i 2 = 1 2 + 2 2 + ⋯ + n 2 = n(n + 1)(2n
6
i=1
n
∑ i3 = 13 + 23 + ⋯ + n3 = n
2
i=0
(n + 1) 2
4
• Left-Endpoint Approximation
A ≈ L n = f (x 0)Δx + f (x 1)Δx + ⋯ + f (x n − 1)Δx =
• Right-Endpoint Approximation
A ≈ R n = f (x 1)Δx + f (x 2)Δx + ⋯ + f (x n)Δx =
• Definite Integral
n
b
∑
∫ a f (x)dx = n lim
→∞
f ⎛⎝x*i ⎞⎠Δx
i=1
• Properties of the Definite Integral
a
∫ a f (x)dx = 0
∫
a
b
f (x)dx = −∫ f (x)dx
a
b
b
∫a
⎡
⎣
b
b
f (x) + g(x)⎤⎦dx = ∫ f (x)dx + ∫ g(x)dx
a
a
⌠ f (x) − g(x) dx = ⌠ f (x)dx − ∫ bg(x)dx
a
⌡a
⌡a
b
⎡
⎣
b
⎤
⎦
b
b
∫ a c f (x)dx = c∫ a f (x)
b
c
for constant c
b
∫ a f (x)dx = ∫ a f (x)dx + ∫ c
f (x)dx
• Mean Value Theorem for Integrals
n
∑
i=1
n
∑
i=1
f (x i − 1)Δx
f (x i)Δx
618
Chapter 5 | Integration
f (x) is continuous over an interval
If
f (c) =
then there is at least one point c ∈ ⎡⎣a, b⎤⎦ such that
⎡
⎣
a, b⎤⎦,
b
1
f (x)dx.
b − a∫a
• Fundamental Theorem of Calculus Part 1
x
If f (x) is continuous over an interval ⎡⎣a, b⎤⎦, and the function F(x) is defined by F(x) = ∫ f (t)dt, then
a
F′ (x) = f (x).
• Fundamental Theorem of Calculus Part 2
If f is continuous over the interval ⎡⎣a, b⎤⎦ and F(x) is any antiderivative of f (x), then
• Net Change Theorem
b
F(b) = F(a) + ∫ F '(x)dx or
a
b
∫ a f (x)dx = F(b) − F(a).
b
∫ a F '(x)dx = F(b) − F(a)
• Substitution with Indefinite Integrals
∫ f g(x) g′ (x)dx = ∫ f (u)du = F(u) + C = F g(x) + C
⎡
⎣
⎤
⎦
⎛
⎝
⎞
⎠
• Substitution with Definite Integrals
⌠ f g(x) g '(x)dx = ∫
g(a)
⌡a
b
⎛
⎝
⎞
⎠
g(b)
f (u)du
• Integrals of Exponential Functions
∫ e x dx = e x + C
⌠a x dx
⌡
x
= a +C
lna
• Integration Formulas Involving Logarithmic Functions
∫ x −1 dx = ln|x| + C
∫ ln x dx = x ln x − x + C = x(ln x − 1) + C
x (ln x − 1) + C
∫ log a x dx = lna
• Integrals That Produce Inverse Trigonometric Functions
⌠ du = sin −1 ⎛⎝u ⎞⎠ + C
a
⌡ a2 − u2
⌠ du = 1 tan −1 ⎛⎝u ⎞⎠ + C
a
⌡a 2 + u 2 a
⌠ du
= 1a sec −1 ⎛⎝ua ⎞⎠ + C
⌡u u 2 − a 2
KEY CONCEPTS
5.1 Approximating Areas
n
• The use of sigma (summation) notation of the form
∑ ai
i=1
form.
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is useful for expressing long sums of values in compact
Chapter 5 | Integration
619
• For a continuous function defined over an interval ⎡⎣a, b⎤⎦, the process of dividing the interval into n equal parts,
extending a rectangle to the graph of the function, calculating the areas of the series of rectangles, and then summing
the areas yields an approximation of the area of that region.
a
• The width of each rectangle is Δx = b −
n .
n
• Riemann sums are expressions of the form
∑
i=1
f ⎛⎝x*i ⎞⎠Δx, and can be used to estimate the area under the curve
⎧
y = f (x). Left- and right-endpoint approximations are special kinds of Riemann sums where the values of ⎨⎩x*i
⎫
⎬
⎭
are chosen to be the left or right endpoints of the subintervals, respectively.
⎧
• Riemann sums allow for much flexibility in choosing the set of points ⎨⎩x*i
⎫
⎬
⎭
at which the function is evaluated,
often with an eye to obtaining a lower sum or an upper sum.
5.2 The Definite Integral
• The definite integral can be used to calculate net signed area, which is the area above the x-axis less the area below
the x-axis. Net signed area can be positive, negative, or zero.
• The component parts of the definite integral are the integrand, the variable of integration, and the limits of
integration.
• Continuous functions on a closed interval are integrable. Functions that are not continuous may still be integrable,
depending on the nature of the discontinuities.
• The properties of definite integrals can be used to evaluate integrals.
• The area under the curve of many functions can be calculated using geometric formulas.
• The average value of a function can be calculated using definite integrals.
5.3 The Fundamental Theorem of Calculus
• The Mean Value Theorem for Integrals states that for a continuous function over a closed interval, there is a value c
such that f (c) equals the average value of the function. See The Mean Value Theorem for Integrals.
• The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. See
Fundamental Theorem of Calculus, Part 1.
• The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an
antiderivative of its integrand. The total area under a curve can be found using this formula. See The
Fundamental Theorem of Calculus, Part 2.
5.4 Integration Formulas and the Net Change Theorem
• The net change theorem states that when a quantity changes, the final value equals the initial value plus the integral
of the rate of change. Net change can be a positive number, a negative number, or zero.
• The area under an even function over a symmetric interval can be calculated by doubling the area over the positive
x-axis. For an odd function, the integral over a symmetric interval equals zero, because half the area is negative.
5.5 Substitution
• Substitution is a technique that simplifies the integration of functions that are the result of a chain-rule derivative.
The term ‘substitution’ refers to changing variables or substituting the variable u and du for appropriate expressions
in the integrand.
• When using substitution for a definite integral, we also have to change the limits of integration.
620
Chapter 5 | Integration
5.6 Integrals Involving Exponential and Logarithmic Functions
• Exponential and logarithmic functions arise in many real-world applications, especially those involving growth and
decay.
• Substitution is often used to evaluate integrals involving exponential functions or logarithms.
5.7 Integrals Resulting in Inverse Trigonometric Functions
• Formulas for derivatives of inverse trigonometric functions developed in Derivatives of Exponential and
Logarithmic Functions lead directly to integration formulas involving inverse trigonometric functions.
• Use the formulas listed in the rule on integration formulas resulting in inverse trigonometric functions to match up
the correct format and make alterations as necessary to solve the problem.
• Substitution is often required to put the integrand in the correct form.
CHAPTER 5 REVIEW EXERCISES
True or False. Justify your answer with a proof or a
counterexample. Assume all functions f and g are
447.
∫
continuous over their domains.
439. If f (x) > 0, f ′ (x) > 0 for all x, then the righthand rule underestimates the integral
b
∫ a f (x). Use a graph
440.
441.
b
∫ a f (x)
If
2
b
b
dx = ∫ f (x)dx∫ f (x)dx
− 2x 2 + 4x⎞⎠dx
4
449.
∫
450.
∫
π/2
2sec(2θ)tan(2θ)dθ
π/3
a
a
f (x) ≤ g(x)
⎛ 3
⎝x
−1
3t dt
448. ⌠
⌡0 1 + 6t 2
to justify your answer.
b
1
for
all
⎡
⎣
⎤
⎦
x ∈ a, b ,
then
b
π/4
2
e cos xsin x cosdx
0
∫ a f (x) ≤ ∫ a g(x).
Find the antiderivative.
442. All continuous functions have an antiderivative.
451.
⌠ dx
⌡(x + 4) 3
452.
∫ x ln⎛⎝x 2⎞⎠dx
453.
⌠ 4x 2 dx
⌡ 1 − x6
Evaluate the Riemann sums L 4 and R 4 for the following
functions over the specified interval. Compare your answer
with the exact answer, when possible, or use a calculator to
determine the answer.
443. y = 3x 2 − 2x + 1 over [−1, 1]
444. y = ln⎛⎝x 2 + 1⎞⎠ over [0, e]
445. y = x 2 sin x over [0, π]
446. y = x + 1x over [1, 4]
Evaluate the following integrals.
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2x
454. ⌠ e 4x dx
⌡1 + e
Find the derivative.
t
455.
d ⌠ sin x dx
dt ⌡ 1 + x 2
0
Chapter 5 | Integration
x3
456.
d
dx ∫ 1
457.
d
dx ∫ 1
458.
d
dx ∫ 0
621
4 − t 2dt
ln(x)
4t + e t⎞⎠dt
⎛
⎝
cos x
2
e t dt
The following problems consider the historic average cost
per gigabyte of RAM on a computer.
Year
5-Year Change ($)
1980
0
1985
−5,468,750
1990
−755,495
1995
−73,005
2000
−29,768
2005
−918
2010
−177
459. If the average cost per gigabyte of RAM in 2010 is
$12, find the average cost per gigabyte of RAM in 1980.
460. The average cost per gigabyte of RAM can be
approximated
by
the
function
C(t) = 8, 500, 000(0.65) t, where t is measured in years
since 1980, and C is cost in US$. Find the average cost per
gigabyte of RAM for 1980 to 2010.
461. Find the average cost of 1GB RAM for 2005 to
2010.
462. The velocity of a bullet from a rifle can be
approximated by v(t) = 6400t 2 − 6505t + 2686, where
t is seconds after the shot and v is the velocity measured
in feet per second. This equation only models the velocity
for the first half-second after the shot: 0 ≤ t ≤ 0.5. What
is the total distance the bullet travels in 0.5 sec?
463. What is the average velocity of the bullet for the first
half-second?
622
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Chapter 5 | Integration
Chapter 6 | Applications of Integration
623
6 | APPLICATIONS OF
INTEGRATION
Figure 6.1 Hoover Dam is one of the United States’ iconic landmarks, and provides irrigation and hydroelectric power for
millions of people in the southwest United States. (credit: modification of work by Lynn Betts, Wikimedia)
Chapter Outline
6.1 Areas between Curves
6.2 Determining Volumes by Slicing
6.3 Volumes of Revolution: Cylindrical Shells
6.4 Arc Length of a Curve and Surface Area
6.5 Physical Applications
6.6 Moments and Centers of Mass
6.7 Integrals, Exponential Functions, and Logarithms
6.8 Exponential Growth and Decay
6.9 Calculus of the Hyperbolic Functions
624
Chapter 6 | Applications of Integration
Introduction
The Hoover Dam is an engineering marvel. When Lake Mead, the reservoir behind the dam, is full, the dam withstands a
great deal of force. However, water levels in the lake vary considerably as a result of droughts and varying water demands.
Later in this chapter, we use definite integrals to calculate the force exerted on the dam when the reservoir is full and we
examine how changing water levels affect that force (see Example 6.28).
Hydrostatic force is only one of the many applications of definite integrals we explore in this chapter. From geometric
applications such as surface area and volume, to physical applications such as mass and work, to growth and decay models,
definite integrals are a powerful tool to help us understand and model the world around us.
6.1 | Areas between Curves
Learning Objectives
6.1.1 Determine the area of a region between two curves by integrating with respect to the
independent variable.
6.1.2 Find the area of a compound region.
6.1.3 Determine the area of a region between two curves by integrating with respect to the
dependent variable.
In Introduction to Integration, we developed the concept of the definite integral to calculate the area below a curve on
a given interval. In this section, we expand that idea to calculate the area of more complex regions. We start by finding the
area between two curves that are functions of x, beginning with the simple case in which one function value is always
greater than the other. We then look at cases when the graphs of the functions cross. Last, we consider how to calculate the
area between two curves that are functions of y.
Area of a Region between Two Curves
Let f (x) and g(x) be continuous functions over an interval ⎡⎣a, b⎤⎦ such that f (x) ≥ g(x) on ⎡⎣a, b⎤⎦. We want to find the
area between the graphs of the functions, as shown in the following figure.
Figure 6.2 The area between the graphs of two functions,
f (x) and g(x), on the interval [a, b].
As we did before, we are going to partition the interval on the x-axis and approximate the area between the graphs
of the functions with rectangles. So, for i = 0, 1, 2,…, n, let P = {x i} be a regular partition of ⎡⎣a, b⎤⎦. Then, for
i = 1, 2,…, n, choose a point x*i ∈ [x i − 1, x i], and on each interval [x i − 1, x i] construct a rectangle that extends
vertically from g(x*i ) to f (x*i ). Figure 6.3(a) shows the rectangles when x*i is selected to be the left endpoint of the
interval and n = 10. Figure 6.3(b) shows a representative rectangle in detail.
Use this calculator (http://www.openstaxcollege.org/l/20_CurveCalc) to learn more about the areas
between two curves.
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Chapter 6 | Applications of Integration
625
Figure 6.3 (a)We can approximate the area between the
graphs of two functions, f (x) and g(x), with rectangles. (b)
The area of a typical rectangle goes from one curve to the other.
The height of each individual rectangle is f (x*i ) − g(x*i ) and the width of each rectangle is Δx. Adding the areas of all
the rectangles, we see that the area between the curves is approximated by
A≈
n
∑ ⎡⎣ f (x*i ) − g(x*i )⎤⎦Δx.
i=1
This is a Riemann sum, so we take the limit as n → ∞ and we get
n
A = n lim
∑ ⎡⎣ f (x*i ) − g(x*i )⎤⎦Δx = ∫
→∞
i=1
b
a
⎡
⎣
f (x) − g(x)⎤⎦dx.
These findings are summarized in the following theorem.
Theorem 6.1: Finding the Area between Two Curves
Let f (x) and g(x) be continuous functions such that f (x) ≥ g(x) over an interval ⎡⎣a, b⎤⎦. Let R denote the region
bounded above by the graph of f (x), below by the graph of g(x), and on the left and right by the lines x = a and
x = b, respectively. Then, the area of R is given by
A=∫
b
⎡
⎣
a
f (x) − g(x)⎤⎦dx.
(6.1)
We apply this theorem in the following example.
Example 6.1
Finding the Area of a Region between Two Curves 1
If R is the region bounded above by the graph of the function f (x) = x + 4 and below by the graph of the
function g(x) = 3 − x over the interval [1, 4], find the area of region R.
2
Solution
The region is depicted in the following figure.
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Chapter 6 | Applications of Integration
Figure 6.4 A region between two curves is shown where one
curve is always greater than the other.
We have
A =∫
b
⎡
⎣
a
4
f (x) − g(x)⎤⎦dx
4
⎤
⎡
⎡
⎞⎤
⎛
= ∫ ⎣(x + 4) − ⎝3 − x ⎠⎦dx = ∫ ⎣3x + 1⎦dx
2
1
1 2
⎡
2
⎤
= ⎣3x + x⎦
4
The area of the region is 57 units 2.
|
4
1
⎛
⎞
= ⎝16 − 7 ⎠ = 57 .
4
4
4
If R is the region bounded by the graphs of the functions f (x) = x + 5 and g(x) = x + 1 over the
2
2
⎡
⎤
interval ⎣1, 5⎦, find the area of region R.
6.1
In Example 6.1, we defined the interval of interest as part of the problem statement. Quite often, though, we want to define
our interval of interest based on where the graphs of the two functions intersect. This is illustrated in the following example.
Example 6.2
Finding the Area of a Region between Two Curves 2
If R is the region bounded above by the graph of the function f (x) = 9 − (x/2) 2 and below by the graph of the
function g(x) = 6 − x, find the area of region R.
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Chapter 6 | Applications of Integration
627
Solution
The region is depicted in the following figure.
Figure 6.5 This graph shows the region below the graph of
f (x) and above the graph of g(x).
We first need to compute where the graphs of the functions intersect. Setting f (x) = g(x), we get
f (x) = g(x)
⎛ ⎞
9 − ⎝x ⎠
2
2
= 6−x
2
9− x = 6−x
4
36 − x 2 = 24 − 4x
x 2 − 4x − 12 = 0
(x − 6)(x + 2) = 0.
The graphs of the functions intersect when x = 6 or x = −2, so we want to integrate from −2 to 6. Since
f (x) ≥ g(x) for −2 ≤ x ≤ 6, we obtain
A =∫
=∫
b
⎡
⎣
f (x) − g(x)⎤⎦dx
a
6⎡
2
⎤
⎢9 − ⎛ x ⎞ − (6 − x)⎥dx =
⎝
⎠
2
⎦
−2⎣
3
2⎤
⎡
= ⎣3x − x + x ⎦
12 2
The area of the region is 64/3 units2.
6.2
|
6
−2
∫
2
⎡
⎤
3 − x + x⎦dx
⎣
4
−2
6
= 64 .
3
If R is the region bounded above by the graph of the function f (x) = x and below by the graph of the
function g(x) = x 4, find the area of region R.
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Chapter 6 | Applications of Integration
Areas of Compound Regions
So far, we have required f (x) ≥ g(x) over the entire interval of interest, but what if we want to look at regions bounded by
the graphs of functions that cross one another? In that case, we modify the process we just developed by using the absolute
value function.
Theorem 6.2: Finding the Area of a Region between Curves That Cross
Let f (x) and g(x) be continuous functions over an interval ⎡⎣a, b⎤⎦. Let R denote the region between the graphs of
f (x) and g(x), and be bounded on the left and right by the lines x = a and x = b, respectively. Then, the area of
R is given by
A=∫
b
| f (x) − g(x)|dx.
a
In practice, applying this theorem requires us to break up the interval ⎡⎣a, b⎤⎦ and evaluate several integrals, depending on
which of the function values is greater over a given part of the interval. We study this process in the following example.
Example 6.3
Finding the Area of a Region Bounded by Functions That Cross
If R is the region between the graphs of the functions f (x) = sin x and g(x) = cos x over the interval [0, π],
find the area of region R.
Solution
The region is depicted in the following figure.
Figure 6.6 The region between two curves can be broken into
two sub-regions.
The graphs of the functions intersect at x = π/4. For x ∈ [0, π/4],
cos x ≥ sin x, so
| f (x) − g(x)| = |sin x − cos x| = cos x − sin x.
On the other hand, for x ∈ [π/4, π],
sin x ≥ cos x, so
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Chapter 6 | Applications of Integration
629
| f (x) − g(x)| = |sin x − cos x| = sin x − cos x.
Then
A =∫
b
| f (x) − g(x)|dx
a
π
π/4
π
0
0
π/4
= ∫ |sin x − cos x|dx = ∫
(cos x − sin x)dx + ∫
π
= [sin x + cos x] | π/4
0 + [−cos x − sin x] | π/4
(sin x − cos x)dx
= ( 2 − 1) + ⎛⎝1 + 2⎞⎠ = 2 2.
The area of the region is 2 2 units2.
6.3
If R is the region between the graphs of the functions f (x) = sin x and g(x) = cos x over the interval
[π/2, 2π], find the area of region R.
Example 6.4
Finding the Area of a Complex Region
Consider the region depicted in Figure 6.7. Find the area of R.
Figure 6.7 Two integrals are required to calculate the area of
this region.
Solution
As with Example 6.3, we need to divide the interval into two pieces. The graphs of the functions intersect at
x = 1 (set f (x) = g(x) and solve for x), so we evaluate two separate integrals: one over the interval [0, 1] and
one over the interval [1, 2].
Over the interval [0, 1], the region is bounded above by f (x) = x 2 and below by the x-axis, so we have
1
3
A 1 = ∫ x 2 dx = x
3
0
|
1
0
= 1.
3
Over the interval [1, 2], the region is bounded above by g(x) = 2 − x and below by the x-axis, so we have
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Chapter 6 | Applications of Integration
2⎤
⎡
A 2 = ∫ (2 − x)dx = ⎣2x − x ⎦
2
1
2
Adding these areas together, we obtain
|
2
1
= 1.
2
A = A1 + A2 = 1 + 1 = 5.
3 2 6
The area of the region is 5/6 units2.
6.4
Consider the region depicted in the following figure. Find the area of R.
Regions Defined with Respect to y
In Example 6.4, we had to evaluate two separate integrals to calculate the area of the region. However, there is another
approach that requires only one integral. What if we treat the curves as functions of y, instead of as functions of x ?
Review Figure 6.7. Note that the left graph, shown in red, is represented by the function y = f (x) = x 2. We could just
as easily solve this for x and represent the curve by the function x = v(y) = y. (Note that x = − y is also a valid
representation of the function y = f (x) = x 2 as a function of y. However, based on the graph, it is clear we are interested
in the positive square root.) Similarly, the right graph is represented by the function y = g(x) = 2 − x, but could just as
easily be represented by the function x = u(y) = 2 − y. When the graphs are represented as functions of y, we see the
region is bounded on the left by the graph of one function and on the right by the graph of the other function. Therefore, if
we integrate with respect to y, we need to evaluate one integral only. Let’s develop a formula for this type of integration.
Let u(y) and v(y) be continuous functions over an interval ⎡⎣c, d⎤⎦ such that u(y) ≥ v(y) for all y ∈ ⎡⎣c, d⎤⎦. We want to
find the area between the graphs of the functions, as shown in the following figure.
Figure 6.8 We can find the area between the graphs of two
functions, u(y) and v(y).
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Chapter 6 | Applications of Integration
631
This time, we are going to partition the interval on the y-axis and use horizontal rectangles to approximate the area between
the functions. So, for i = 0, 1, 2,…, n, let Q = {y i} be a regular partition of ⎡⎣c, d⎤⎦. Then, for i = 1, 2,…, n, choose
a point y*i ∈ [y i − 1, y i], then over each interval [y i − 1, y i] construct a rectangle that extends horizontally from v⎛⎝y*i
⎞
⎠
to u⎛⎝y*i ⎞⎠. Figure 6.9(a) shows the rectangles when y*i is selected to be the lower endpoint of the interval and n = 10.
Figure 6.9(b) shows a representative rectangle in detail.
Figure 6.9 (a) Approximating the area between the graphs of
two functions, u(y) and v(y), with rectangles. (b) The area of
a typical rectangle.
The height of each individual rectangle is Δy and the width of each rectangle is u⎛⎝y*i ⎞⎠ − v⎛⎝y*i ⎞⎠. Therefore, the area
between the curves is approximately
A≈
n
∑ ⎡⎣u⎛⎝y*i ⎞⎠ − v⎛⎝y*i ⎞⎠⎤⎦Δy.
i=1
This is a Riemann sum, so we take the limit as n → ∞, obtaining
n
d
A = n lim
∑ ⎡⎣u⎛⎝y*i ⎞⎠ − v⎛⎝y*i ⎞⎠⎤⎦Δy = ∫ ⎡⎣u(y) − v(y)⎤⎦dy.
→∞
c
i=1
These findings are summarized in the following theorem.
Theorem 6.3: Finding the Area between Two Curves, Integrating along the y-axis
Let u(y) and v(y) be continuous functions such that u(y) ≥ v(y) for all y ∈ ⎡⎣c, d⎤⎦. Let R denote the region bounded
on the right by the graph of u(y), on the left by the graph of v(y), and above and below by the lines y = d and
y = c, respectively. Then, the area of R is given by
d
A = ∫ ⎡⎣u(y) − v(y)⎤⎦dy.
c
Example 6.5
Integrating with Respect to y
(6.2)
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Chapter 6 | Applications of Integration
Let’s revisit Example 6.4, only this time let’s integrate with respect to y. Let R be the region depicted in
Figure 6.10. Find the area of R by integrating with respect to y.
Figure 6.10 The area of region R can be calculated using
one integral only when the curves are treated as functions of y.
Solution
We must first express the graphs as functions of y. As we saw at the beginning of this section, the curve on
the left can be represented by the function x = v(y) = y, and the curve on the right can be represented by the
function x = u(y) = 2 − y.
Now we have to determine the limits of integration. The region is bounded below by the x-axis, so the lower limit
of integration is y = 0. The upper limit of integration is determined by the point where the two graphs intersect,
which is the point (1, 1), so the upper limit of integration is y = 1. Thus, we have ⎡⎣c, d⎤⎦ = [0, 1].
Calculating the area of the region, we get
d
A = ∫ ⎡⎣u(y) − v(y)⎤⎦dy
c
=∫
1
0
⎡
2 − y⎞⎠ − y⎤⎦dy = ⎢ 2y −
⎡⎛
⎣⎝
= 5.
6
⎣
y 2 2 3/2⎤⎥
− y
2 3
⎦
|
1
0
The area of the region is 5/6 units2.
6.5 Let’s revisit the checkpoint associated with Example 6.4, only this time, let’s integrate with respect to
y. Let be the region depicted in the following figure. Find the area of R by integrating with respect to y.
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Chapter 6 | Applications of Integration
633
6.1 EXERCISES
For the following exercises, determine the area of the
region between the two curves in the given figure by
integrating over the x-axis.
4. y = cos θ and y = 0.5, for 0 ≤ θ ≤ π
1. y = x 2 − 3 and y = 1
For the following exercises, determine the area of the
region between the two curves by integrating over the
y-axis.
5. x = y 2 and x = 9
2. y = x 2 and y = 3x + 4
6. y = x and x = y 2
For the following exercises, split the region between the
two curves into two smaller regions, then determine the
area by integrating over the x-axis. Note that you will
have two integrals to solve.
3. y = x 3 and y = x 2 + x
For the following exercises, graph the equations and shade
the area of the region between the curves. Determine its
area by integrating over the x-axis.
7. y = x 2 and y = −x 2 + 18x
8. y = 1x , y = 12 , and x = 3
x
9. y = cos x and y = cos 2 x on x = [−π, π]
10. y = e x, y = e 2x − 1, and x = 0
11. y = e x, y = e −x, x = −1 and x = 1
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Chapter 6 | Applications of Integration
12. y = e, y = e x, and y = e −x
31. y = 4 − 3x and y = 1x
13. y = |x| and y = x 2
32. y = sin x, x = −π/6, x = π/6, and y = cos 3 x
For the following exercises, graph the equations and shade
the area of the region between the curves. If necessary,
break the region into sub-regions to determine its entire
area.
33. y = x 2 − 3x + 2 and y = x 3 − 2x 2 − x + 2
14. y = sin(πx), y = 2x, and x > 0
15. y = 12 − x, y = x, and y = 1
16. y = sin x and y = cos x over x = [−π, π]
3
2
17. y = x and y = x − 2x over x = [−1, 1]
2
18. y = x + 9 and y = 10 + 2x over x = [−1, 3]
19. y = x 3 + 3x and y = 4x
For the following exercises, graph the equations and shade
the area of the region between the curves. Determine its
area by integrating over the y-axis.
20. x = y 3 and x = 3y − 2
21. x = 2y and x = y 3 − y
22. x = −3 + y 2 and x = y − y 2
23. y 2 = x and x = y + 2
24. x = |y| and 2x = −y 2 + 2
25. x = sin y, x = cos(2y), y = π/2, and y = −π/2
For the following exercises, graph the equations and shade
the area of the region between the curves. Determine its
area by integrating over the x-axis or y-axis, whichever
seems more convenient.
26. x = y 4 and x = y 5
27. y = xe x, y = e x, x = 0, and x = 1
28. y = x 6 and y = x 4
29. x = y 3 + 2y 2 + 1 and x = −y 2 + 1
30. y = |x| and y = x 2 − 1
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34. y = 2 cos 3 (3x), y = −1, x = π , and x = − π
4
4
35. y + y 3 = x and 2y = x
36. y = 1 − x 2 and y = x 2 − 1
37. y = cos −1 x, y = sin −1 x, x = −1, and x = 1
For the following exercises, find the exact area of the
region bounded by the given equations if possible. If you
are unable to determine the intersection points analytically,
use a calculator to approximate the intersection points with
three decimal places and determine the approximate area of
the region.
38. [T] x = e y and y = x − 2
39. [T] y = x 2 and y = 1 − x 2
40. [T] y = 3x 2 + 8x + 9 and 3y = x + 24
41. [T] x = 4 − y 2 and y 2 = 1 + x 2
42. [T] x 2 = y 3 and x = 3y
43.
y = sin 3 x + 2, y = tan x, x = −1.5, and x = 1.5
44. [T] y = 1 − x 2 and y 2 = x 2
45. [T] y = 1 − x 2 and y = x 2 + 2x + 1
46. [T] x = 4 − y 2 and x = 1 + 3y + y 2
47. [T] y = cos x, y = e x, x = −π, and x = 0
[T]
Chapter 6 | Applications of Integration
635
48. The largest triangle with a base on the x-axis that
2
2
fits inside the upper half of the unit circle y + x = 1
is given by y = 1 + x and y = 1 − x. See the following
figure. What is the area inside the semicircle but outside the
triangle?
obtain the same answer?
53. y = x 2 + 2x + 1 and y = −x 2 − 3x + 4
54. y = x 4 and x = y 5
55. x = y 2 − 2 and x = 2y
For the following exercises, solve using calculus, then
check your answer with geometry.
56. Determine the equations for the sides of the square
that touches the unit circle on all four sides, as seen in the
following figure. Find the area between the perimeter of
this square and the unit circle. Is there another way to solve
this without using calculus?
49. A factory selling cell phones has a marginal cost
function C(x) = 0.01x 2 − 3x + 229, where x represents
the number of cell phones, and a marginal revenue function
given by R(x) = 429 − 2x. Find the area between the
graphs of these curves and x = 0. What does this area
represent?
50.
An amusement park has a marginal cost function
C(x) = 1000e −x + 5, where x represents the number
of tickets sold, and a marginal revenue function given by
R(x) = 60 − 0.1x. Find the total profit generated when
selling 550 tickets. Use a calculator to determine
intersection points, if necessary, to two decimal places.
51. The tortoise versus the hare: The speed of the hare
is given by the sinusoidal function H(t) = 1 − cos⎛⎝(πt)/2⎞⎠
whereas
the
speed
T(t) = (1/2)tan −1 (t/4),
of
the
tortoise
is
where t is time measured in
hours and the speed is measured in miles per hour. Find the
area between the curves from time t = 0 to the first time
after one hour when the tortoise and hare are traveling at
the same speed. What does it represent? Use a calculator to
determine the intersection points, if necessary, accurate to
three decimal places.
52. The tortoise versus the hare: The speed of the hare
is
given
by
the
sinusoidal
function
H(t) = (1/2) − (1/2)cos(2πt) whereas the speed of the
tortoise is T(t) = t, where t is time measured in hours
and speed is measured in kilometers per hour. If the race is
over in 1 hour, who won the race and by how much? Use a
calculator to determine the intersection points, if necessary,
accurate to three decimal places.
For the following exercises, find the area between the
curves by integrating with respect to x and then with
respect to y. Is one method easier than the other? Do you
57. Find the area between the perimeter of the unit circle
and the triangle created from y = 2x + 1, y = 1 − 2x and
y = − 3 , as seen in the following figure. Is there a way
5
to solve this without using calculus?
636
Chapter 6 | Applications of Integration
6.2 | Determining Volumes by Slicing
Learning Objectives
6.2.1 Determine the volume of a solid by integrating a cross-section (the slicing method).
6.2.2 Find the volume of a solid of revolution using the disk method.
6.2.3 Find the volume of a solid of revolution with a cavity using the washer method.
In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite
integrals to find volumes of three-dimensional solids. We consider three approaches—slicing, disks, and washers—for
finding these volumes, depending on the characteristics of the solid.
Volume and the Slicing Method
Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional
solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid,
for example, can be computed by multiplying length, width, and height: V = lwh. The formulas for the volume of a sphere
⎛
4 3⎞
⎝V = 3 πr ⎠,
⎛
⎞
⎛
⎞
a cone ⎝V = 1 πr 2 h⎠, and a pyramid ⎝V = 1 Ah⎠ have also been introduced. Although some of these
3
3
formulas were derived using geometry alone, all these formulas can be obtained by using integration.
We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as
a soup can or a metal rod, in mathematics the word cylinder has a more general meaning. To discuss cylinders in this more
general context, we first need to define some vocabulary.
We define the cross-section of a solid to be the intersection of a plane with the solid. A cylinder is defined as any solid
that can be generated by translating a plane region along a line perpendicular to the region, called the axis of the cylinder.
Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure 6.11 is an example
of a cylinder with a noncircular base. To calculate the volume of a cylinder, then, we simply multiply the area of the crosssection by the height of the cylinder: V = A · h. In the case of a right circular cylinder (soup can), this becomes V = πr 2 h.
Figure 6.11 Each cross-section of a particular cylinder is identical to the others.
If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for
its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid
into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be
parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the
solid S shown in Figure 6.12, extending along the x-axis.
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Chapter 6 | Applications of Integration
637
Figure 6.12 A solid with a varying cross-section.
We want to divide S into slices perpendicular to the x-axis. As we see later in the chapter, there may be times when we
want to slice the solid in some other direction—say, with slices perpendicular to the y-axis. The decision of which way to
slice the solid is very important. If we make the wrong choice, the computations can get quite messy. Later in the chapter,
we examine some of these situations in detail and look at how to decide which way to slice the solid. For the purposes of
this section, however, we use slices perpendicular to the x-axis.
Because the cross-sectional area is not constant, we let A(x) represent the area of the cross-section at point x. Now let
⎧
⎨
⎩
⎫
⎬
⎭
P = x 0, x 1 …, X n be a regular partition of ⎡⎣a, b⎤⎦, and for i = 1, 2,…n, let S i represent the slice of S stretching from
x i − 1 to x i. The following figure shows the sliced solid with n = 3.
Figure 6.13 The solid S has been divided into three slices
perpendicular to the x-axis.
Finally, for i = 1, 2,…n, let x*i be an arbitrary point in [x i − 1, x i]. Then the volume of slice S i can be estimated by
V ⎛⎝S i⎞⎠ ≈ A⎛⎝x*i ⎞⎠Δx. Adding these approximations together, we see the volume of the entire solid S can be approximated by
V(S) ≈
n
∑
i=1
A⎛⎝x*i ⎞⎠Δx.
By now, we can recognize this as a Riemann sum, and our next step is to take the limit as n → ∞. Then we have
n
b
i=1
a
V(S) = n lim
∑ A⎛⎝x*i ⎞⎠Δx = ∫ A(x)dx.
→∞
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Chapter 6 | Applications of Integration
The technique we have just described is called the slicing method. To apply it, we use the following strategy.
Problem-Solving Strategy: Finding Volumes by the Slicing Method
1. Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture
if one is not provided.
2. Determine a formula for the area of the cross-section.
3. Integrate the area formula over the appropriate interval to get the volume.
Recall that in this section, we assume the slices are perpendicular to the x-axis. Therefore, the area formula is in terms of
x and the limits of integration lie on the x-axis. However, the problem-solving strategy shown here is valid regardless of
how we choose to slice the solid.
Example 6.6
Deriving the Formula for the Volume of a Pyramid
We know from geometry that the formula for the volume of a pyramid is V = 1 Ah. If the pyramid has a square
3
base, this becomes V = 1 a 2 h, where a denotes the length of one side of the base. We are going to use the
3
slicing method to derive this formula.
Solution
We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid
shown in Figure 6.14, oriented along the x-axis.
Figure 6.14 (a) A pyramid with a square base is oriented along the x-axis. (b) A two-dimensional view of the
pyramid is seen from the side.
We first want to determine the shape of a cross-section of the pyramid. We are know the base is a square, so the
cross-sections are squares as well (step 1). Now we want to determine a formula for the area of one of these crosssectional squares. Looking at Figure 6.14(b), and using a proportion, since these are similar triangles, we have
s x
ax
a = h or s = h .
Therefore, the area of one of the cross-sectional squares is
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639
⎛
⎞
2
A(x) = s 2 = ⎝ax ⎠ ⎛⎝step 2⎞⎠.
h
Then we find the volume of the pyramid by integrating from 0 to h (step 3):
h
V = ∫ A(x)dx
0
h
=∫
0
⎛ax ⎞
a2
⎝ h ⎠ dx = 2
h
2
⎡ 2 ⎛ ⎞⎤
= a 2 ⎝1 x 3⎠
⎣h 3 ⎦
This is the formula we were looking for.
6.6
|
h
0
h
∫ x 2dx
0
= 1 a 2 h.
3
Use the slicing method to derive the formula V = 1 πr 2 h for the volume of a circular cone.
3
Solids of Revolution
If a region in a plane is revolved around a line in that plane, the resulting solid is called a solid of revolution, as shown in
the following figure.
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Chapter 6 | Applications of Integration
Figure 6.15 (a) This is the region that is revolved around the x-axis.
(b) As the region begins to revolve around the axis, it sweeps out a
solid of revolution. (c) This is the solid that results when the
revolution is complete.
Solids of revolution are common in mechanical applications, such as machine parts produced by a lathe. We spend the rest
of this section looking at solids of this type. The next example uses the slicing method to calculate the volume of a solid of
revolution.
Use an online integral calculator (http://www.openstaxcollege.org/l/20_IntCalc2) to learn more.
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Example 6.7
Using the Slicing Method to find the Volume of a Solid of Revolution
Use the slicing method to find the volume of the solid of revolution bounded by the graphs of
f (x) = x 2 − 4x + 5, x = 1, and x = 4, and rotated about the x-axis.
Solution
Using the problem-solving strategy, we first sketch the graph of the quadratic function over the interval [1, 4] as
shown in the following figure.
Figure 6.16 A region used to produce a solid of revolution.
Next, revolve the region around the x-axis, as shown in the following figure.
Figure 6.17 Two views, (a) and (b), of the solid of revolution produced by revolving the region
in Figure 6.16 about the x-axis.
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Chapter 6 | Applications of Integration
Since the solid was formed by revolving the region around the x-axis, the cross-sections are circles (step 1).
The area of the cross-section, then, is the area of a circle, and the radius of the circle is given by f (x). Use the
formula for the area of the circle:
2
A(x) = πr 2 = π ⎡⎣ f (x)⎤⎦ 2 = π ⎛⎝x 2 − 4x + 5⎞⎠ (step 2).
The volume, then, is (step 3)
h
V = ∫ A(x)dx
a
4
2
= ∫ π ⎛⎝x 2 − 4x + 5⎞⎠ dx = π ∫
1
4
1
⎛ 4
⎝x
− 8x 3 + 26x 2 − 40x + 25⎞⎠dx
|
3
⎞
⎛ 5
= π ⎝x − 2x 4 + 26x − 20x 2 + 25x⎠ 1 = 78 π.
3
5
5
The volume is 78π/5.
4
6.7 Use the method of slicing to find the volume of the solid of revolution formed by revolving the region
between the graph of the function f (x) = 1/x and the x-axis over the interval [1, 2] around the x-axis. See
the following figure.
The Disk Method
When we use the slicing method with solids of revolution, it is often called the disk method because, for solids of
revolution, the slices used to over approximate the volume of the solid are disks. To see this, consider the solid of revolution
generated by revolving the region between the graph of the function f (x) = (x − 1) 2 + 1 and the x-axis over the interval
[−1, 3] around the x-axis. The graph of the function and a representative disk are shown in Figure 6.18(a) and (b). The
region of revolution and the resulting solid are shown in Figure 6.18(c) and (d).
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643
Figure 6.18 (a) A thin rectangle for approximating the area under a curve. (b) A representative disk formed by
revolving the rectangle about the x-axis. (c) The region under the curve is revolved about the x-axis, resulting in
(d) the solid of revolution.
We already used the formal Riemann sum development of the volume formula when we developed the slicing method. We
know that
b
V = ∫ A(x)dx.
a
The only difference with the disk method is that we know the formula for the cross-sectional area ahead of time; it is the
area of a circle. This gives the following rule.
Rule: The Disk Method
Let f (x) be continuous and nonnegative. Define R as the region bounded above by the graph of f (x), below by the
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Chapter 6 | Applications of Integration
x-axis, on the left by the line x = a, and on the right by the line x = b. Then, the volume of the solid of revolution
formed by revolving R around the x-axis is given by
b
(6.3)
V = ∫ π ⎡⎣ f (x)⎤⎦ 2 dx.
a
The volume of the solid we have been studying (Figure 6.18) is given by
b
V = ∫ π ⎡⎣ f (x)⎤⎦ 2 dx
=∫
a
3
2
π ⎡⎣(x − 1) 2 + 1⎤⎦ dx = π ∫
−1
3
⎡
⎣(x
−1
3
2
− 1) 4 + 2(x − 1) 2 + 1⎤⎦ dx
|
⎤
⎡
⎡⎛
⎞ ⎛
⎞⎤
= π ⎣1 (x − 1) 5 + 2 (x − 1) 3 + x⎦ −1 = π ⎣⎝32 + 16 + 3⎠ − ⎝− 32 − 16 − 1⎠⎦ = 412π units 3.
3
3
3
5
5
15
5
Let’s look at some examples.
Example 6.8
Using the Disk Method to Find the Volume of a Solid of Revolution 1
Use the disk method to find the volume of the solid of revolution generated by rotating the region between the
graph of f (x) = x and the x-axis over the interval [1, 4] around the x-axis.
Solution
The graphs of the function and the solid of revolution are shown in the following figure.
Figure 6.19 (a) The function f (x) = x over the interval [1, 4]. (b) The solid of revolution
obtained by revolving the region under the graph of f (x) about the x-axis.
We have
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Chapter 6 | Applications of Integration
645
b
V = ∫ π ⎡⎣ f (x)⎤⎦ 2 dx
a
4
4
= ∫ π[ x] 2dx = π ∫ x dx
1
|
4
1
= π x 2 1 = 15π .
2
2
The volume is (15π)/2 units .
3
6.8 Use the disk method to find the volume of the solid of revolution generated by rotating the region
between the graph of f (x) = 4 − x and the x-axis over the interval [0, 4] around the x-axis.
So far, our examples have all concerned regions revolved around the x-axis, but we can generate a solid of revolution by
revolving a plane region around any horizontal or vertical line. In the next example, we look at a solid of revolution that has
been generated by revolving a region around the y-axis. The mechanics of the disk method are nearly the same as when
the x-axis is the axis of revolution, but we express the function in terms of y and we integrate with respect to y as well.
This is summarized in the following rule.
Rule: The Disk Method for Solids of Revolution around the y-axis
Let g(y) be continuous and nonnegative. Define Q as the region bounded on the right by the graph of g(y), on the
left by the y-axis, below by the line y = c, and above by the line y = d. Then, the volume of the solid of revolution
formed by revolving Q around the y-axis is given by
d
V = ∫ π ⎡⎣g(y)⎤⎦ 2 dy.
(6.4)
c
The next example shows how this rule works in practice.
Example 6.9
Using the Disk Method to Find the Volume of a Solid of Revolution 2
Let R be the region bounded by the graph of g(y) = 4 − y and the y-axis over the y-axis interval [0, 4].
Use the disk method to find the volume of the solid of revolution generated by rotating R around the y-axis.
Solution
Figure 6.20 shows the function and a representative disk that can be used to estimate the volume. Notice that
since we are revolving the function around the y-axis, the disks are horizontal, rather than vertical.
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Chapter 6 | Applications of Integration
Figure 6.20 (a) Shown is a thin rectangle between the curve of the function g(y) = 4 − y
and the y-axis. (b) The rectangle forms a representative disk after revolution around the y-axis.
The region to be revolved and the full solid of revolution are depicted in the following figure.
Figure 6.21 (a) The region to the left of the function g(y) = 4 − y over the y-axis interval
[0, 4]. (b) The solid of revolution formed by revolving the region about the y-axis.
To find the volume, we integrate with respect to y. We obtain
d
V = ∫ π ⎡⎣g(y)⎤⎦ 2 dy
c
4
4
= ∫ π ⎡⎣ 4 − y⎤⎦ 2 dy = π ∫ ⎛⎝4 − y⎞⎠dy
0
|
4
⎡
y 2 ⎤⎥
⎢
= π 4y −
= 8π.
2⎦ 0
⎣
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Chapter 6 | Applications of Integration
647
The volume is 8π units3.
6.9 Use the disk method to find the volume of the solid of revolution generated by rotating the region
between the graph of g(y) = y and the y-axis over the interval [1, 4] around the y-axis.
The Washer Method
Some solids of revolution have cavities in the middle; they are not solid all the way to the axis of revolution. Sometimes,
this is just a result of the way the region of revolution is shaped with respect to the axis of revolution. In other cases, cavities
arise when the region of revolution is defined as the region between the graphs of two functions. A third way this can happen
is when an axis of revolution other than the x-axis or y-axis is selected.
When the solid of revolution has a cavity in the middle, the slices used to approximate the volume are not disks, but washers
(disks with holes in the center). For example, consider the region bounded above by the graph of the function f (x) = x
and below by the graph of the function g(x) = 1 over the interval [1, 4]. When this region is revolved around the x-axis,
the result is a solid with a cavity in the middle, and the slices are washers. The graph of the function and a representative
washer are shown in Figure 6.22(a) and (b). The region of revolution and the resulting solid are shown in Figure 6.22(c)
and (d).
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Chapter 6 | Applications of Integration
Figure 6.22 (a) A thin rectangle in the region between two curves. (b) A
representative disk formed by revolving the rectangle about the x-axis. (c) The region
between the curves over the given interval. (d) The resulting solid of revolution.
The cross-sectional area, then, is the area of the outer circle less the area of the inner circle. In this case,
A(x) = π( x) 2 − π(1) 2 = π(x − 1).
Then the volume of the solid is
b
V = ∫ A(x)dx
a
4
⎡
2
⎤
= ∫ π(x − 1)dx = π ⎣ x − x⎦
2
1
Generalizing this process gives the washer method.
|
4
1
= 9 π units 3.
2
Rule: The Washer Method
Suppose f (x) and g(x) are continuous, nonnegative functions such that f (x) ≥ g(x) over ⎡⎣a, b⎤⎦. Let R denote the
region bounded above by the graph of f (x), below by the graph of g(x), on the left by the line x = a, and on
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Chapter 6 | Applications of Integration
649
the right by the line x = b. Then, the volume of the solid of revolution formed by revolving R around the x-axis is
given by
b
(6.5)
V = ∫ π ⎡⎣⎛⎝ f (x)⎞⎠ 2 − ⎛⎝g(x)⎞⎠ 2⎤⎦dx.
a
Example 6.10
Using the Washer Method
Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of f (x) = x
and below by the graph of g(x) = 1/x over the interval [1, 4] around the x-axis.
Solution
The graphs of the functions and the solid of revolution are shown in the following figure.
Figure 6.23 (a) The region between the graphs of the functions f (x) = x and
g(x) = 1/x over the interval [1, 4]. (b) Revolving the region about the x-axis generates
a solid of revolution with a cavity in the middle.
We have
b
V = ∫ π ⎡⎣⎛⎝ f (x)⎞⎠ 2 − ⎛⎝g(x)⎞⎠ 2⎤⎦dx
a
= π∫
4⎡
⎤
⎡ 3
⎤
x 2 − ⎛⎝1x ⎞⎠ dx = π ⎣ x + 1x ⎦
⎦
3
1⎣
2
|
4
1
= 81π units 3.
4
6.10 Find the volume of a solid of revolution formed by revolving the region bounded by the graphs of
f (x) = x and g(x) = 1/x over the interval [1, 3] around the x-axis.
As with the disk method, we can also apply the washer method to solids of revolution that result from revolving a region
around the y-axis. In this case, the following rule applies.
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Chapter 6 | Applications of Integration
Rule: The Washer Method for Solids of Revolution around the y-axis
Suppose u(y) and v(y) are continuous, nonnegative functions such that v(y) ≤ u(y) for y ∈ ⎡⎣c, d⎤⎦. Let Q denote
the region bounded on the right by the graph of u(y), on the left by the graph of v(y), below by the line y = c,
and above by the line y = d. Then, the volume of the solid of revolution formed by revolving Q around the y-axis
is given by
d
V = ∫ π ⎡⎣⎛⎝u(y)⎞⎠ 2 − ⎛⎝v(y)⎞⎠ 2⎤⎦dy.
c
Rather than looking at an example of the washer method with the y-axis as the axis of revolution, we now consider an
example in which the axis of revolution is a line other than one of the two coordinate axes. The same general method
applies, but you may have to visualize just how to describe the cross-sectional area of the volume.
Example 6.11
The Washer Method with a Different Axis of Revolution
Find the volume of a solid of revolution formed by revolving the region bounded above by f (x) = 4 − x and
below by the x-axis over the interval [0, 4] around the line y = −2.
Solution
The graph of the region and the solid of revolution are shown in the following figure.
Figure 6.24 (a) The region between the graph of the function f (x) = 4 − x and the x-axis
over the interval [0, 4]. (b) Revolving the region about the line y = −2 generates a solid of
revolution with a cylindrical hole through its middle.
We can’t apply the volume formula to this problem directly because the axis of revolution is not one of the
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651
coordinate axes. However, we still know that the area of the cross-section is the area of the outer circle less the
area of the inner circle. Looking at the graph of the function, we see the radius of the outer circle is given by
f (x) + 2, which simplifies to
f (x) + 2 = (4 − x) + 2 = 6 − x.
The radius of the inner circle is g(x) = 2. Therefore, we have
4
V = ∫ π ⎡⎣(6 − x) 2 − (2) 2⎤⎦dx
0
= π∫
4
0
⎛ 2
⎝x
⎤
⎡ 3
− 12x + 32⎞⎠dx = π ⎣ x − 6x 2 + 32x⎦
3
|
4
0
= 160π units 3.
3
6.11 Find the volume of a solid of revolution formed by revolving the region bounded above by the graph of
f (x) = x + 2 and below by the x-axis over the interval [0, 3] around the line y = −1.
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6.2 EXERCISES
58. Derive the formula for the volume of a sphere using
the slicing method.
59. Use the slicing method to derive the formula for the
volume of a cone.
66. A pyramid with height 5 units, and an isosceles
triangular base with lengths of 6 units and 8 units, as seen
here.
60. Use the slicing method to derive the formula for the
volume of a tetrahedron with side length a.
61. Use the disk method to derive the formula for the
volume of a trapezoidal cylinder.
62. Explain when you would use the disk method versus
the washer method. When are they interchangeable?
For the following exercises, draw a typical slice and find
the volume using the slicing method for the given volume.
63. A pyramid with height 6 units and square base of side
2 units, as pictured here.
67. A cone of radius r and height h has a smaller cone of
radius r/2 and height h/2 removed from the top, as seen
here. The resulting solid is called a frustum.
64. A pyramid with height 4 units and a rectangular base
with length 2 units and width 3 units, as pictured here.
For the following exercises, draw an outline of the solid and
find the volume using the slicing method.
68. The base is a circle of radius a. The slices
perpendicular to the base are squares.
69. The base is a triangle with vertices (0, 0), (1, 0),
65. A tetrahedron with a base side of 4 units, as seen here.
and (0, 1). Slices perpendicular to the xy-plane are
semicircles.
70. The base is the region under the parabola y = 1 − x 2
in the first quadrant. Slices perpendicular to the xy-plane
are squares.
71. The base is the region under the parabola y = 1 − x 2
and above the x-axis. Slices perpendicular to the y-axis
are squares.
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Chapter 6 | Applications of Integration
72.
The base is the region enclosed by y = x 2 and
y = 9. Slices perpendicular to the x-axis are right isosceles
653
90. y = x + 2, y = x + 6, x = 0, and x = 5
triangles.
91. y = x 2 and y = x + 2
73. The base is the area between y = x and y = x 2.
92. x 2 = y 3 and x 3 = y 2
Slices perpendicular to the x-axis are semicircles.
For the following exercises, draw the region bounded by
the curves. Then, use the disk method to find the volume
when the region is rotated around the x-axis.
93. y = 4 − x 2 and y = 2 − x
94. [T] y = cos x, y = e −x, x = 0, and x = 1.2927
74. x + y = 8, x = 0, and y = 0
95. y = x and y = x 2
75. y = 2x 2, x = 0, x = 4, and y = 0
96. y = sin x, y = 5 sin x, x = 0 and x = π
76. y = e x + 1, x = 0, x = 1, and y = 0
97. y = 1 + x 2 and y = 4 − x 2
77. y = x 4, x = 0, and y = 1
For the following exercises, draw the region bounded by
the curves. Then, use the washer method to find the volume
when the region is revolved around the y-axis.
78. y = x, x = 0, x = 4, and y = 0
79. y = sin x, y = cos x, and x = 0
80. y = 1x , x = 2, and y = 3
2
2
81. x − y = 9 and x + y = 9, y = 0 and x = 0
For the following exercises, draw the region bounded by
the curves. Then, find the volume when the region is
rotated around the y-axis.
82. y = 4 − 1 x, x = 0, and y = 0
2
83. y = 2x 3, x = 0, x = 1, and y = 0
84. y = 3x 2, x = 0, and y = 3
85. y = 4 − x 2, y = 0, and x = 0
86. y =
1 , x = 0, and x = 3
x+1
87. x = sec(y) and y = π , y = 0 and x = 0
4
88. y =
1 , x = 0, and x = 2
x+1
89. y = 4 − x, y = x, and x = 0
For the following exercises, draw the region bounded by
the curves. Then, find the volume when the region is
rotated around the x-axis.
98. y = x, x = 4, and y = 0
99. y = x + 2, y = 2x − 1, and x = 0
100. y = 3 x and y = x 3
2y
101. x = e , x = y 2, y = 0, and y = ln(2)
102. x = 9 − y 2, x = e
−y
, y = 0, and y = 3
654
103.
Chapter 6 | Applications of Integration
Yogurt containers can be shaped like frustums.
1 x around the y-axis to find the
Rotate the line y = m
volume between y = a and y = b.
106. A better approximation of the volume of a football
is given by the solid that comes from rotating y = sin x
around the x-axis from x = 0 to x = π. What is the
volume of this football approximation, as seen here?
107. What is the volume of the Bundt cake that comes
from rotating y = sin x around the y-axis from x = 0 to
x = π?
104. Rotate the ellipse ⎛⎝x 2 /a 2⎞⎠ + ⎛⎝y 2 /b 2⎞⎠ = 1 around the
x-axis to approximate the volume of a football, as seen
here.
105. Rotate the ellipse ⎛⎝x 2 /a 2⎞⎠ + ⎛⎝y 2 /b 2⎞⎠ = 1 around the
y-axis to approximate the volume of a football.
For the following exercises, find the volume of the solid
described.
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Chapter 6 | Applications of Integration
108. The base is the region between y = x and y = x 2.
Slices perpendicular to the x-axis are semicircles.
109. The base is the region enclosed by the generic ellipse
⎛ 2 2⎞
⎛ 2 2⎞
⎝x /a ⎠ + ⎝y /b ⎠ = 1. Slices perpendicular to the x-axis
are semicircles.
110. Bore a hole of radius a down the axis of a right cone
and through the base of radius b, as seen here.
111. Find the volume common to two spheres of radius r
with centers that are 2h apart, as shown here.
112. Find the volume of a spherical cap of height h and
radius r where h < r, as seen here.
655
113. Find the volume of a sphere of radius R with a cap
of height h removed from the top, as seen here.
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Chapter 6 | Applications of Integration
6.3 | Volumes of Revolution: Cylindrical Shells
Learning Objectives
6.3.1 Calculate the volume of a solid of revolution by using the method of cylindrical shells.
6.3.2 Compare the different methods for calculating a volume of revolution.
In this section, we examine the method of cylindrical shells, the final method for finding the volume of a solid of revolution.
We can use this method on the same kinds of solids as the disk method or the washer method; however, with the disk and
washer methods, we integrate along the coordinate axis parallel to the axis of revolution. With the method of cylindrical
shells, we integrate along the coordinate axis perpendicular to the axis of revolution. The ability to choose which variable
of integration we want to use can be a significant advantage with more complicated functions. Also, the specific geometry
of the solid sometimes makes the method of using cylindrical shells more appealing than using the washer method. In the
last part of this section, we review all the methods for finding volume that we have studied and lay out some guidelines to
help you determine which method to use in a given situation.
The Method of Cylindrical Shells
Again, we are working with a solid of revolution. As before, we define a region R, bounded above by the graph of a
function y = f (x), below by the x-axis, and on the left and right by the lines x = a and x = b, respectively, as shown
in Figure 6.25(a). We then revolve this region around the y-axis, as shown in Figure 6.25(b). Note that this is different
from what we have done before. Previously, regions defined in terms of functions of x were revolved around the x-axis
or a line parallel to it.
Figure 6.25 (a) A region bounded by the graph of a function of x. (b) The solid of revolution formed when the
region is revolved around the y-axis.
As we have done many times before, partition the interval ⎡⎣a, b⎤⎦ using a regular partition, P = {x 0, x 1 ,…, x n} and,
for i = 1, 2,…, n, choose a point x*i ∈ [x i − 1, x i]. Then, construct a rectangle over the interval [x i − 1, x i] of height
f (x*i ) and width Δx. A representative rectangle is shown in Figure 6.26(a). When that rectangle is revolved around the
y-axis, instead of a disk or a washer, we get a cylindrical shell, as shown in the following figure.
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Chapter 6 | Applications of Integration
657
Figure 6.26 (a) A representative rectangle. (b) When this rectangle is revolved around the y-axis, the result is a cylindrical
shell. (c) When we put all the shells together, we get an approximation of the original solid.
To calculate the volume of this shell, consider Figure 6.27.
Figure 6.27 Calculating the volume of the shell.
The shell is a cylinder, so its volume is the cross-sectional area multiplied by the height of the cylinder. The cross-sections
are annuli (ring-shaped regions—essentially, circles with a hole in the center), with outer radius x i and inner radius x i − 1.
Thus, the cross-sectional area is πx i2 − πx i2− 1. The height of the cylinder is f (x*i ). Then the volume of the shell is
V shell = f (x*i )(πx i2 − πx i2− 1)
= π f (x*i )⎛⎝x i2 − x i2− 1⎞⎠
= π f (x*i )(x i + x i − 1)(x i − x i − 1)
⎛x + x i − 1 ⎞
= 2π f (x*i )⎝ i
⎠(x i − x i − 1).
2
Note that x i − x i − 1 = Δx, so we have
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Chapter 6 | Applications of Integration
⎛x i + x i − 1 ⎞
⎠Δx.
2
V shell = 2π f (x*i )⎝
xi + xi − 1
is both the midpoint of the interval [x i − 1, x i] and the average radius of the shell, and we can
2
approximate this by x*i . We then have
Furthermore,
V shell ≈ 2π f (x*i )x*i Δx.
Another way to think of this is to think of making a vertical cut in the shell and then opening it up to form a flat plate
(Figure 6.28).
Figure 6.28 (a) Make a vertical cut in a representative shell. (b) Open the shell up to form a flat plate.
In reality, the outer radius of the shell is greater than the inner radius, and hence the back edge of the plate would be slightly
longer than the front edge of the plate. However, we can approximate the flattened shell by a flat plate of height f (x*i ),
width 2πx*i , and thickness Δx (Figure 6.28). The volume of the shell, then, is approximately the volume of the flat
plate. Multiplying the height, width, and depth of the plate, we get
V shell ≈ f (x*i )⎛⎝2πx*i ⎞⎠Δx,
which is the same formula we had before.
To calculate the volume of the entire solid, we then add the volumes of all the shells and obtain
V≈
n
∑ ⎛⎝2πx*i
i=1
f (x*i )Δx⎞⎠.
Here we have another Riemann sum, this time for the function 2πx f (x). Taking the limit as n → ∞ gives us
n
b
V = n lim
∑ ⎛⎝2πx*i f (x*i )Δx⎞⎠ = ∫ ⎛⎝2πx f (x)⎞⎠dx.
→∞
i=1
a
This leads to the following rule for the method of cylindrical shells.
Rule: The Method of Cylindrical Shells
Let f (x) be continuous and nonnegative. Define R as the region bounded above by the graph of f (x), below by the
x-axis, on the left by the line x = a, and on the right by the line x = b. Then the volume of the solid of revolution
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Chapter 6 | Applications of Integration
659
formed by revolving R around the y-axis is given by
b
V = ∫ ⎛⎝2πx f (x)⎞⎠dx.
(6.6)
a
Now let’s consider an example.
Example 6.12
The Method of Cylindrical Shells 1
Define R as the region bounded above by the graph of f (x) = 1/x and below by the x-axis over the interval
[1, 3]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
Solution
First we must graph the region R and the associated solid of revolution, as shown in the following figure.
Figure 6.29 (a) The region R under the graph of f (x) = 1/x over the
interval [1, 3]. (b) The solid of revolution generated by revolving R about
the y-axis.
Then the volume of the solid is given by
b
V = ∫ ⎛⎝2πx f (x)⎞⎠dx
=∫
a
3
⎛
⎛1 ⎞⎞
⎝2πx⎝ x ⎠⎠dx
1
3
= ∫ 2π dx = 2πx| 31 = 4π units 3 .
1
6.12
Define R as the region bounded above by the graph of f (x) = x 2 and below by the x-axis over the
interval [1, 2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
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Chapter 6 | Applications of Integration
Example 6.13
The Method of Cylindrical Shells 2
Define R as the region bounded above by the graph of f (x) = 2x − x 2 and below by the x-axis over the interval
[0, 2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
Solution
First graph the region R and the associated solid of revolution, as shown in the following figure.
Figure 6.30 (a) The region R under the graph of f (x) = 2x − x 2 over
the interval [0, 2]. (b) The volume of revolution obtained by revolving
R about the y-axis.
Then the volume of the solid is given by
b
V = ∫ ⎛⎝2πx f (x)⎞⎠dx
a
2
= ∫ ⎛⎝2πx⎛⎝2x − x 2⎞⎠⎞⎠dx = 2π ∫
0
4⎤
⎡ 3
= 2π ⎣2x − x ⎦
3
4
6.13
|
2
0
2
0
⎛ 2
⎝2x
− x 3⎞⎠dx
= 8π units 3 .
3
Define R as the region bounded above by the graph of f (x) = 3x − x 2 and below by the x-axis over
the interval [0, 2]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
As with the disk method and the washer method, we can use the method of cylindrical shells with solids of revolution,
revolved around the x-axis, when we want to integrate with respect to y. The analogous rule for this type of solid is given
here.
Rule: The Method of Cylindrical Shells for Solids of Revolution around the x-axis
Let g(y) be continuous and nonnegative. Define Q as the region bounded on the right by the graph of g(y), on
the left by the y-axis, below by the line y = c, and above by the line y = d. Then, the volume of the solid of
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661
revolution formed by revolving Q around the x-axis is given by
d
V = ∫ ⎛⎝2πyg(y)⎞⎠dy.
c
Example 6.14
The Method of Cylindrical Shells for a Solid Revolved around the x-axis
Define Q as the region bounded on the right by the graph of g(y) = 2 y and on the left by the y-axis for
y ∈ [0, 4]. Find the volume of the solid of revolution formed by revolving Q around the x-axis.
Solution
First, we need to graph the region Q and the associated solid of revolution, as shown in the following figure.
Figure 6.31 (a) The region Q to the left of the function g(y) over the interval
[0, 4]. (b) The solid of revolution generated by revolving Q around the x-axis.
Label the shaded region Q. Then the volume of the solid is given by
d
V = ∫ ⎛⎝2πyg(y)⎞⎠dy
c
4
4
= ∫ ⎛⎝2πy⎛⎝2 y⎞⎠⎞⎠dy = 4π ∫ y 3/2dy
0
|
0
⎡ 2y 5/2 ⎤ 4
⎥ = 256π units 3 .
= 4π ⎢
5
⎣ 5 ⎦ 0
6.14
Define Q as the region bounded on the right by the graph of g(y) = 3/y and on the left by the y-axis
for y ∈ [1, 3]. Find the volume of the solid of revolution formed by revolving Q around the x-axis.
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Chapter 6 | Applications of Integration
For the next example, we look at a solid of revolution for which the graph of a function is revolved around a line other than
one of the two coordinate axes. To set this up, we need to revisit the development of the method of cylindrical shells. Recall
that we found the volume of one of the shells to be given by
V shell = f (x*i )(πx i2 − πx i2− 1)
= π f (x*i )⎛⎝x i2 − x i2− 1⎞⎠
= π f (x*i )(x i + x i − 1)(x i − x i − 1)
⎛x + x i − 1 ⎞
= 2π f (x*i )⎝ i
⎠(x i − x i − 1).
2
This was based on a shell with an outer radius of x i and an inner radius of x i − 1. If, however, we rotate the region around
a line other than the y-axis, we have a different outer and inner radius. Suppose, for example, that we rotate the region
around the line x = −k, where k is some positive constant. Then, the outer radius of the shell is x i + k and the inner
radius of the shell is x i − 1 + k. Substituting these terms into the expression for volume, we see that when a plane region is
rotated around the line x = −k, the volume of a shell is given by
⎛⎛⎝x i + k⎞⎠ + ⎛⎝x i − 1 + k⎞⎠⎞⎛⎛
⎛
⎞⎞
⎞
⎠⎝⎝x i + k⎠ − ⎝x i − 1 + k⎠⎠
⎝
2
V shell = 2π f (x*i )
⎛⎛ x i + x i − 2 ⎞
⎞
⎠ + k⎠Δx.
2
= 2π f (x*i )⎝⎝
As before, we notice that
xi + xi − 1
is the midpoint of the interval [x i − 1, x i] and can be approximated by x*i . Then,
2
the approximate volume of the shell is
V shell ≈ 2π ⎛⎝x*i + k⎞⎠ f (x*i )Δx.
The remainder of the development proceeds as before, and we see that
b
V = ∫ ⎛⎝2π(x + k) f (x)⎞⎠dx.
a
We could also rotate the region around other horizontal or vertical lines, such as a vertical line in the right half plane. In
each case, the volume formula must be adjusted accordingly. Specifically, the x-term in the integral must be replaced with
an expression representing the radius of a shell. To see how this works, consider the following example.
Example 6.15
A Region of Revolution Revolved around a Line
Define R as the region bounded above by the graph of f (x) = x and below by the x-axis over the interval
[1, 2]. Find the volume of the solid of revolution formed by revolving R around the line x = −1.
Solution
First, graph the region R and the associated solid of revolution, as shown in the following figure.
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663
Figure 6.32 (a) The region R between the graph of f (x) and the x-axis over the interval [1, 2]. (b) The
solid of revolution generated by revolving R around the line x = −1.
Note that the radius of a shell is given by x + 1. Then the volume of the solid is given by
2
V = ∫ ⎛⎝2π(x + 1) f (x)⎞⎠dx
1
2
= ∫ (2π(x + 1)x)dx = 2π ∫
1
2⎤
⎡ 3
= 2π ⎣ x + x ⎦
3
2
6.15
|
2
1
2
1
⎛ 2
⎝x
+ x⎞⎠dx
= 23π units 3 .
3
Define R as the region bounded above by the graph of f (x) = x 2 and below by the x-axis over the
interval [0, 1]. Find the volume of the solid of revolution formed by revolving R around the line x = −2.
For our final example in this section, let’s look at the volume of a solid of revolution for which the region of revolution is
bounded by the graphs of two functions.
Example 6.16
A Region of Revolution Bounded by the Graphs of Two Functions
Define R as the region bounded above by the graph of the function f (x) = x and below by the graph of the
function g(x) = 1/x over the interval [1, 4]. Find the volume of the solid of revolution generated by revolving
R around the y-axis.
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Chapter 6 | Applications of Integration
Solution
First, graph the region R and the associated solid of revolution, as shown in the following figure.
Figure 6.33 (a) The region R between the graph of f (x) and the graph of g(x) over the interval [1, 4]. (b)
The solid of revolution generated by revolving R around the y-axis.
Note that the axis of revolution is the y-axis, so the radius of a shell is given simply by x. We don’t need to
make any adjustments to the x-term of our integrand. The height of a shell, though, is given by f (x) − g(x), so
in this case we need to adjust the f (x) term of the integrand. Then the volume of the solid is given by
4
V = ∫ ⎛⎝2πx⎛⎝ f (x) − g(x)⎞⎠⎞⎠dx
=∫
1
4
1
⎛
⎛
⎝2πx⎝
⎡
5/2
= 2π ⎣2x
5
6.16
x − 1x ⎞⎠⎞⎠dx = 2π ∫
⎤
− x⎦
|
4
1
4
1
⎛ 3/2
⎝x
− 1⎞⎠dx
= 94π units 3.
5
Define R as the region bounded above by the graph of f (x) = x and below by the graph of g(x) = x 2
over the interval [0, 1]. Find the volume of the solid of revolution formed by revolving R around the y-axis.
Which Method Should We Use?
We have studied several methods for finding the volume of a solid of revolution, but how do we know which method to use?
It often comes down to a choice of which integral is easiest to evaluate. Figure 6.34 describes the different approaches
for solids of revolution around the x-axis. It’s up to you to develop the analogous table for solids of revolution around the
y-axis.
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665
Figure 6.34
Let’s take a look at a couple of additional problems and decide on the best approach to take for solving them.
Example 6.17
Selecting the Best Method
For each of the following problems, select the best method to find the volume of a solid of revolution generated
by revolving the given region around the x-axis, and set up the integral to find the volume (do not evaluate the
integral).
a. The region bounded by the graphs of y = x,
y = 2 − x, and the x-axis.
b. The region bounded by the graphs of y = 4x − x 2 and the x-axis.
Solution
a. First, sketch the region and the solid of revolution as shown.
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Chapter 6 | Applications of Integration
Figure 6.35 (a) The region R bounded by two lines and the x-axis. (b) The solid of
revolution generated by revolving R about the x-axis.
Looking at the region, if we want to integrate with respect to x, we would have to break the integral
into two pieces, because we have different functions bounding the region over [0, 1] and [1, 2]. In this
case, using the disk method, we would have
1
2
V = ∫ ⎛⎝πx 2⎞⎠dx + ∫ ⎛⎝π(2 − x) 2⎞⎠dx.
0
1
If we used the shell method instead, we would use functions of y to represent the curves, producing
1
V = ∫ ⎛⎝2πy⎡⎣⎛⎝2 − y⎞⎠ − y⎤⎦⎞⎠dy
0
1
= ∫ ⎛⎝2πy⎡⎣2 − 2y⎤⎦⎞⎠dy.
0
Neither of these integrals is particularly onerous, but since the shell method requires only one integral,
and the integrand requires less simplification, we should probably go with the shell method in this case.
b. First, sketch the region and the solid of revolution as shown.
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Figure 6.36 (a) The region R between the curve and the x-axis. (b) The solid of
revolution generated by revolving R about the x-axis.
Looking at the region, it would be problematic to define a horizontal rectangle; the region is bounded on
the left and right by the same function. Therefore, we can dismiss the method of shells. The solid has no
cavity in the middle, so we can use the method of disks. Then
4
2
V = ∫ π ⎛⎝4x − x 2⎞⎠ dx.
0
6.17 Select the best method to find the volume of a solid of revolution generated by revolving the given
region around the x-axis, and set up the integral to find the volume (do not evaluate the integral): the region
bounded by the graphs of y = 2 − x 2 and y = x 2.
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Chapter 6 | Applications of Integration
6.3 EXERCISES
For the following exercise, find the volume generated when
the region between the two curves is rotated around the
given axis. Use both the shell method and the washer
method. Use technology to graph the functions and draw a
typical slice by hand.
114. [T] Over the curve of y = 3x, x = 0, and y = 3
129. y = 5x 3 − 2x 4, x = 0, and x = 2
For the following exercises, use shells to find the volume
generated by rotating the regions between the given curve
and y = 0 around the x-axis.
rotated around the y-axis.
130. y = 1 − x 2, x = 0, and x = 1
115. [T] Under the curve of y = 3x, x = 0, and x = 3
131. y = x 2, x = 0, and x = 2
rotated around the y-axis.
116. [T] Over the curve of y = 3x, x = 0, and y = 3
132. y = e x, x = 0, and x = 1
rotated around the x-axis.
133. y = ln(x), x = 1, and x = e
117. [T] Under the curve of y = 3x, x = 0, and x = 3
134. x =
1 , y = 1, and y = 4
1 + y2
135. x =
1 + y2
y , y = 0, and y = 2
rotated around the x-axis.
118. [T] Under the curve of y = 2x 3, x = 0, and x = 2
rotated around the y-axis.
3
119. [T]